Question: As shown in the figure, in the Cartesian coordinate system, there are points A and B on the graph of the inverse proportion function $y=\frac{k}{x} (k>0, x>0)$. Their $x$-coordinates are 2 and 4, respectively. If the area of $\triangle ABO$ is 6, what is the value of $k$?

Solution:
\textbf{Solution:} As shown in the figure, draw $AC \perp x$-axis at point $C$, and draw $BD \perp x$-axis at point $D$,\\
since the $x$-coordinates of points $A$ and $B$ are 2 and 4, respectively,\\
therefore $A\left(2, \frac{k}{2}\right)$, $B\left(4, \frac{k}{4}\right)$, and the area of $\triangle AOC$ and $\triangle BOD$ are both $\frac{1}{2}|k|$,\\
since the area of $\triangle AOB$ equals the area of $\triangle AOC$ plus the area of trapezoid $ACDB$ minus the area of $\triangle BOD$,\\
thus the area of $\triangle AOB$ equals the area of trapezoid $ACDB$ which is 6,\\
thus $\frac{1}{2}\left(\frac{k}{2}+\frac{k}{4}\right)\times (4-2)=6$,\\
solving for $k$ gives $k=\boxed{8}$.


Question: As shown in the figure, the graph of the inverse proportion function $y=\frac{k}{x}$ ($k \neq 0$) passes through the intersection point P of the diagonals of rectangle $ABCD$. It is known that points A, C, and D are on the coordinate axes, $BD \perp DC$, and the area of rectangle $ABCD$ is 8. What is the value of $k$?

Solution:
\textbf{Solution:} Draw $PE\perp y$-axis at point E.\\
Since quadrilateral ABCD is a parallelogram,\\
it follows that AB=CD.\\
Also, since BD$\perp x$-axis,\\
quadrilateral ABDO is a rectangle.\\
Hence, AB=DO.\\
Therefore, the area of rectangle ABDO, which is also the area of parallelogram ABCD, is 8.\\
Since P is the intersection point of the diagonals and $PE\perp y$-axis,\\
quadrilateral PDOE is a rectangle with area 4.\\
That is, $DO\cdot EO=4$.\\
Let the coordinates of point P be $(x,y)$.\\
Then $k=x\boxed{y=-4}$.


Question: Place a set of triangle rulers as shown in the diagram. If $\angle 1 = 27^\circ 40'$, what is the measure of $\angle 2$?

Solution:
\textbf{Solution:} Given, $\because$ $\angle 1=27^\circ40'$ and $\angle$BAC=$60^\circ$, \\
$\therefore$ $\angle$EAC=$\angle$BAC−$\angle 1 = 60^\circ−27^\circ40'=32^\circ20'$, \\
$\because$ $\angle$DAE=$90^\circ$, \\
$\therefore$ $\angle 2=\angle$DAE−$\angle$EAC=$90^\circ−32^\circ20'=\boxed{57^\circ40'}$.


Question: As shown in the figure, $BC= \frac{1}{2} AB$, and $D$ is the midpoint of $AC$. If $DB=1$, what is the length of $AB$?

Solution:
\textbf{Solution:} Given $\because BC= \frac{1}{2} AB$,\\
$\therefore AC=AB+BC= \frac{3}{2}AB$,\\
$\because D$ is the midpoint of $AC$,\\
$\therefore CD= \frac{1}{2}AC=\frac{1}{2}\times \frac{3}{2}AB=\frac{3}{4}AB$,\\
$\therefore DB=CD-BC= \frac{3}{4}AB-\frac{1}{2}AB=\frac{1}{4}AB$,\\
which means $\frac{1}{4}AB=1$,\\
$\therefore AB=\boxed{4}$.


Question: As shown in the figure, point C is a point on the line segment $AB$, point M is the midpoint of line segment $AC$, and point N is the midpoint of line segment $BC$. Given that $AB=20$, what is the length of $MN$?

Solution:
\textbf{Solution:} Let $M$ be the midpoint of segment $AC$, and $N$ be the midpoint of segment $BC$,\\
$\therefore$ $MC=\frac{1}{2}AC$, $NC=\frac{1}{2}BC$,\\
$\therefore$ $MN=MC+NC=\frac{1}{2}\left(AC+BC\right)=\frac{1}{2}AB$,\\
Given that $AB=20$,\\
$\therefore$ $MN=\frac{1}{2}\times 20=\boxed{10}$.


Question: As shown in the figure, it is known that the line segment $AB=4\,m$. The line segment $AB$ is extended to point $C$, such that $BC=2AB$. If point $D$ is the midpoint of line segment $AC$, how many $cm$ is the line segment $BD$?

Solution:
\textbf{Solution:} Given $\because$AB=4, BC=2AB, \\
$\therefore$BC=$2\times 4$=8, \\
$\therefore$AC=AB+BC=4+8=12, \\
$\because$ point D is the midpoint of segment AC, \\
$\therefore$AD=$\frac{1}{2}$AC=6, \\
$\therefore$BD=AD−AB=6−4=\boxed{2}


Question: As shown in the figure, on a foldable number line, points $A$ and $B$ represent the numbers $-8$ and $3$, respectively. If point $C$ is used as the folding point and the number line is folded to the right, such that point $A$ falls to the right of point $B$ and the distance between points $A$ and $B$ is $1$ unit, what is the number represented by point $C$?

Solution:
\textbf{Solution:} Given that points A and B represent the numbers -8 and 3, respectively, \\
$\therefore$ $AB=3-\left(-8\right)=11$, \\
Given that the number line is folded to the right at point C, if point A falls on the right side of point B, and points A and B are 1 unit length apart, \\
$\therefore$ $BC=\frac{11-1}{2}=5$, \\
$\therefore$ the number represented by point C is: $3-5=\boxed{-2}$.


Question: As shown in the figure, $BC=4\,\text{cm}$, $BD=7\,\text{cm}$, and point $D$ is the midpoint of $AC$. What is the length of $AC$ in centimeters?

Solution:
\textbf{Solution:} Given that $CD = BD - BC = 7 - 4 = 3\,(\text{cm})$\\
Since point D is the midpoint of AC, it follows that $AC = 2CD = \boxed{6}\,(\text{cm})$


Question: As shown in the figure, line AD intersects with BE at point O, angles $\angle DOE$ and $\angle COE$ are complementary. If $\angle COE=72^\circ$, what is the measure of $\angle AOB$?

Solution:
Solution: Since $\angle DOE$ and $\angle COE$ are complementary,\\
therefore $\angle DOE + \angle COE = 90^\circ$,\\
since $\angle COE = 72^\circ$,\\
therefore $\angle DOE = 18^\circ$,\\
thus $\angle AOB = \angle DOE = \boxed{18^\circ}$


Question: As shown in the figure, lines AB and CD intersect at point O, with OE bisecting $\angle AOC$. If $\angle BOD=60^\circ$, what is the measure of $\angle AOE$?

Solution:
\textbf{Solution:} Since $\angle BOD=60^\circ$,\\
it follows that $\angle AOC=\angle BOD=60^\circ$,\\
since OE bisects $\angle AOC$,\\
thus $\angle AOE= \frac{1}{2}\angle AOC=30^\circ$,\\
therefore, the answer is: $\boxed{30^\circ}$.


Question: As shown in the figure, within $\angle AOB$ there are 3 rays $OC$, $OD$, $OE$. If $\angle AOC= 51^\circ$, $\angle BOE = \frac{1}{3}\angle BOC$, $\angle BOD=\frac{1}{3}\angle AOB$, then what is the degree measure of $\angle DOE$?

Solution:
\textbf{Solution:} Let $\angle BOE = x$,\\
since $\angle BOE= \frac{1}{3}\angle BOC$,\\
thus, $\angle BOC=3\angle BOE=3x$,\\
consequently, $\angle AOB=\angle AOC+\angle BOC=51^\circ+3x$.\\
Since $\angle BOD= \frac{1}{3}\angle AOB$,\\
thus, $\angle BOD= \frac{1}{3}\times(51^\circ+3x)=17^\circ+x$,\\
therefore, $\angle DOE=\angle BOD-\angle BOE=17^\circ+x-x=\boxed{17^\circ}$.


Question: As shown in the figure, it is known that the length of segment AB is $a$. Extend segment AB to point C so that $BC = \frac{1}{2}AB$. Take point D as the midpoint of segment AC. If $DB=2$, what is the value of $a$?

Solution:
Solution: Since AB=a, \\
therefore BC= $\frac{1}{2}$a, \\
therefore AC=AB+BC=$\frac{3}{2}a$, \\
since point D is the midpoint of AC, \\
therefore CD=$\frac{1}{2}$AC=$\frac{3}{4}$a, \\
therefore DB=CD−BC=$\frac{3}{4}$a−$\frac{1}{2}$a=2, \\
therefore a=\boxed{8}.


Question: As shown in the figure, the straight lines $AB$ and $CD$ intersect at the point $O$, $OE \perp CD$ with the foot of the perpendicular at point $O$. If $\angle BOE = 40^\circ 6'$, what is the degree measure of $\angle AOC$?

Solution:
Solution: Since $OE \perp CD$, \\
therefore $\angle EOC = 90^\circ$.\\
Since $\angle BOE = 40^\circ6'$, \\
therefore $\angle AOC = 180^\circ - 90^\circ - 40^\circ6' = \boxed{49^\circ54'}$.


Question: As shown in the figure, lines AB and CD intersect at O, OA bisects $\angle COE$, and $\angle COE \colon \angle BOE = 2 \colon 5$, then what is the degree measure of $\angle EOD$?

Solution:
\textbf{Solution:} Since $\angle COE : \angle BOE = 2 : 5$,\\
let $\angle COE = 2x$, then $\angle BOE = 5x$,\\
Since OA bisects $\angle COE$,\\
thus $\angle AOE = \angle COA = \angle BOD = x$,\\
thus $\angle DOE = 4x$,\\
Since $\angle COE + \angle DOE = 180^\circ$,\\
thus $2x + 4x = 180^\circ$,\\
thus $x = 30^\circ$,\\
thus $\angle DOE = 4x = \boxed{120^\circ}$.


Question: As shown in the figure, the lines AB and CD intersect at point O. If $\angle BOD = 40^\circ$ and OA bisects $\angle COE$, what is the measure of $\angle AOE$?

Solution:
\textbf{Solution:} Since $\angle BOD = 40^\circ$,\\
therefore $\angle AOC = \angle BOD = 40^\circ$,\\
since OA bisects $\angle COE$,\\
therefore $\angle AOE = \angle AOC = \boxed{40^\circ}$.


Question: As shown in the diagram, it is known that $AB \parallel CD$, $\angle 1 = \angle 2$, $\angle E = 50^\circ$. What is the measure of $\angle F$?

Solution:
\textbf{Solution:} Draw line BC, \\
$\because$AB$\parallel$CD, \\
$\therefore$$\angle$ABC=$\angle$BCD, \\
$\because$$\angle$1=$\angle$2, \\
$\therefore$$\angle$EBC=$\angle$BCF, \\
$\therefore$EB$\parallel$CF, \\
$\therefore$$\angle$F=$\angle$E=$50^\circ$. \\
Therefore, the answer is: $\boxed{50^\circ}$.


Question: As shown in the figure, in quadrilateral $ABCD$, $\angle DAC=\angle ACB$, $\angle D=86^\circ$, what is the measure of $\angle BCD$ in degrees?

Solution:
\textbf{Solution:} Since $\angle DAC = \angle ACB$, \\
it follows that $AD \parallel BC$, \\
hence $\angle D + \angle BCD = 180^\circ$, \\
since $\angle D = 86^\circ$, \\
it follows that $\angle BCD = 180^\circ - \angle D = 180^\circ - 86^\circ = \boxed{94^\circ}$.


Question: As shown in the figure, line segment $BD = \frac{1}{3}AB = \frac{1}{4}CD$, points $M$ and $N$ are the midpoints of line segments $AB$ and $CD$, respectively, and $MN = 20\,cm$, then what is the length of $AC$ in $cm$?

Solution:
\textbf{Solution:} Since BD $= \frac{1}{3}$AB $= \frac{1}{4}$CD, we have AB=3BD and CD=4BD.\\
Because points M and N are the midpoints of segments AB and CD, respectively,\\
it follows that AM=BM $= \frac{1}{2}AB = \frac{3}{2}$BD and DN=CN $= \frac{1}{2}CD = 2$BD.\\
Using the property of segment addition and subtraction, we find BN=DN−BD=2BD−BD=BD and BC=CD−BD=4BD−BD=3BD,\\
therefore, MN=MB+BN $= \frac{3}{2}$BD+BD=20.\\
Solving gives BD=8cm.\\
Thus, AC=AB+BC=3BD+3BD=6BD=6×8=$\boxed{48}$,


Question: As shown in the figure, it is known that $\angle BAC=130^\circ$, $AB=AC$, and the perpendicular bisector of $AC$ intersects $BC$ at point $D$. What is the degree measure of $\angle BAD$?

Solution:
\textbf{Solution:} Given that DE is the perpendicular bisector of AC,\\
it follows that AD=CD,\\
thus, $\angle C=\angle CAD$,\\
since AB=AC,\\
it also follows that $\angle B=\angle C$,\\
in $\triangle ABC$, $\angle ABC+\angle B+\angle C=180^\circ$,\\
therefore, $130^\circ+2\angle C=180^\circ$,\\
solving this, we find $\angle C=25^\circ$,\\
thus, $\angle ADB=\angle CAD+\angle C=25^\circ+25^\circ=50^\circ$.\\
Therefore, $\angle BAD=180^\circ-\angle B-\angle ADB=\boxed{105^\circ}$,


Question: As shown in the figure, $P$ is a point inside the equilateral triangle $ABC$. When $\triangle ABP$ is rotated $60^\circ$ clockwise around point $B$, it becomes $\triangle CBP'$. If $PB = 3$, what is the length of $PP'$?

Solution:
\textbf{Solution:} We have, since rotating $\triangle ABP$ $60^\circ$ clockwise about point B yields $\triangle CBP'$,\\
thus $\angle PBP'=60^\circ$ and $BP=BP'$,\\
therefore, $\triangle BPP'$ is an equilateral triangle,\\
hence $PP'=BP=\boxed{3}$


Question: As shown in the diagram, a rectangular paper ABCD is folded along MN such that point A lands on point A' on side BC, and the corresponding point for D is D'. Connect A'D' and intersect side CD at point E. Connect CD'. If AB = 9, AD = 6, and point A' is the midpoint of BC, then what is the length of segment ED'?

Solution:
\textbf{Solution:} Fold the rectangle $ABCD$ along $MN$, so that point $A$ falls on the side $BC$ at point $A'$,\\
$\therefore AM=A'M$, $\angle MA'D'=\angle A=90^\circ$,\\
Let $AM=A'M=x$, then $BM=9-x$,\\
$\because A'$ is the midpoint of $BC$,\\
$\therefore A'B=A'C=\frac{1}{2}BC=\frac{1}{2}AD=3$,\\
In $\triangle A'BM$,\\
$A'B^2+BM^2=A'M^2$,\\
that is $3^2+(9-x)^2=x^2$,\\
Solving for $x$ gives: $x=5$,\\
$\therefore A'M=5$, $MB=4$,\\
$\because \angle MA'B+\angle EA'C=90^\circ$, $\angle A'EC+\angle EA'C=90^\circ$,\\
$\therefore \angle MA'B=\angle A'EC$,\\
$\because \angle B=\angle A'CE=90^\circ$,\\
$\therefore \triangle A'BM\sim \triangle ECA'$,\\
$\therefore \frac{A'E}{A'M}=\frac{A'C}{BM}$, that is $\frac{A'E}{5}=\frac{3}{4}$,\\
$\therefore A'E=\frac{15}{4}$,\\
$\therefore ED'=A'D'-A'E=AD-A'E=6-\frac{15}{4}=\boxed{\frac{9}{4}}$


Question: As shown in the figure, $OP$ bisects $\angle AOB$, $\angle AOP=15^\circ$, $PC\parallel OA$, $PC=4$, $PD\perp OA$, with the foot of the perpendicular being $D$. What is the length of $PD$?

Solution:
\textbf{Solution:} Let PE$\perp$OB at E,\\
since $\angle$BOP=$\angle$AOP, PD$\perp$OA, PE$\perp$OB,\\
therefore PE=PD (The distances from a point on the angle bisector to the two sides of the angle are equal),\\
since $\angle$BOP=$\angle$AOP=15$^\circ$,\\
therefore $\angle$AOB=30$^\circ$,\\
since PC$\parallel$OA,\\
therefore $\angle$BCP=$\angle$AOB=30$^\circ$,\\
therefore, in Rt$\triangle$PCE, PE=$\frac{1}{2}$PC=$\frac{1}{2}\times 4=2 $(In a right triangle, the leg opposite the 30$^\circ$ angle is half the length of the hypotenuse),\\
therefore, PD=PE=\boxed{2},


Question: As shown in the figure, in the sector $AOB$, $\angle AOB=90^\circ$, with radius $OA=6$. When the sector $AOB$ is folded along the line passing through point $B$, point $O$ falls exactly on point $D$ on the arc $AB$. What is the area of the entire shaded region? (Express in terms of $\pi$)

Solution:
\textbf{Solution:} Draw $OD$. According to the properties of folding, $CD=CO$, $BD=BO$, $\angle DBC=\angle OBC$,\\
$\therefore OB=OD=BD$, i.e., $\triangle OBD$ is an equilateral triangle,\\
$\therefore \angle DBO=60^\circ$, $\therefore \angle CBO= \frac{1}{2} \angle DBO=30^\circ$,\\
$\because \angle AOB=90^\circ$, $\therefore OC=OB \cdot \tan \angle CBO=6 \times \frac{\sqrt{3}}{3} =2\sqrt{3}$,\\
$\therefore S_{\triangle BDC}=S_{\triangle OBC}= \frac{1}{2} \times OB \times OC= \frac{1}{2} \times 6 \times 2\sqrt{3} =6\sqrt{3}$,\\
$S_{\text{sector} AOB}= \frac{90}{360} \pi \times 6^2=9\pi$,\\
$\therefore$ the area of the entire shaded region is: $S_{\text{sector} AOB}-S_{\triangle BDC}-S_{\triangle OBC}=9\pi-6\sqrt{3}-6\sqrt{3} =9\pi-12\sqrt{3}$.\\
Therefore, the answer is: $9\pi-12\sqrt{3}=\boxed{9\pi-12\sqrt{3}}$.


Question: As shown in the figure, a square ABCD with a side length of 2 rolls clockwise on the outside of a regular hexagon with equal sides. It stops when it returns to the starting position after rolling around once. What is the maximum distance from point A to the starting point during the rolling process?

Solution:
\textbf{Solution:} As shown in the figure, the trajectory of point A is the arc in the diagram. Extend AE to intersect the arc at H, the length of segment AH, which is the maximum distance from point A to the starting point during the rolling process.\\
$\because$ EH＝EA$_{2}$＝$\sqrt{2^2+2^2}=2\sqrt{2}$,\\
In $\triangle$AEF, $\because$AF＝EF＝2, $\angle$AFE＝$120^\circ$,\\
$\therefore$AE＝$2\times AF\times \cos30^\circ=2\sqrt{3}$,\\
$\therefore$AH＝AE+EH＝$2\sqrt{3}+2\sqrt{2}$,\\
$\therefore$The maximum distance from point A to the starting point during the rolling process is $\boxed{2\sqrt{3}+2\sqrt{2}}$.


Question: As shown in the figure, point $O$ is a point on side $AB$ of $\triangle ABC$, where $\angle ACB = 90^\circ$. A circle $\odot O$ with radius OB is drawn and tangent to AC at point D. If $BC = 4$ and $\sin A = \frac{4}{5}$, what is the length of the radius of $\odot O$?

Solution:
\textbf{Solution:} Solve: In $\triangle ABC$ with a right angle at $B$, we have $BC=4$, and $\sin A=\frac{4}{5}$,\\
\(\therefore \frac{BC}{AB}=\frac{4}{5}\), that is, \(\frac{4}{AB}=\frac{4}{5}\),\\
\(\therefore AB=5\),\\
Connect $OD$,\\
\(\because AC\) is a tangent to circle $O$,\\
\(\therefore OD \perp AC\),\\
Let the radius of circle $O$ be $r$, then $OD = OB = r$,\\
\(\therefore AO=5-r\),\\
In $\triangle AOD$ with a right angle at $O$, and \(\sin A = \frac{4}{5}\),\\
\(\therefore \frac{OD}{AO} = \frac{4}{5}\), that is, \(\frac{r}{5-r} = \frac{4}{5}\),\\
\(\therefore r=\boxed{\frac{20}{9}}\).


Question: In the figure, in $ \triangle ABC$, $ \angle C=90^\circ$, $AC=BC=3$. Points $D$ and $E$ are on sides $AC$ and $AB$ respectively. Folding $ \triangle ADE$ along the line $DE$ makes point $A$ land on side $BC$, denoted as point $F$. If $CF=1$, what is the value of $BE$?

Solution:
\textbf{Solution:} As shown in the figure, draw $FG\perp AB$ at point G through point F,

$\because$ in $\triangle ABC$, $\angle C=90^\circ$, $AC=BC=3$,

$\therefore AB=\sqrt{AC^2+BC^2}=3\sqrt{2}$, $\tan B=\frac{AC}{BC}=1$,

$\therefore \angle A=\angle B=45^\circ$,

$\therefore \triangle FGB$ is an isosceles right triangle,

$\therefore BG=FG=FB\cdot \sin B=\sqrt{2}$,

Let $BE=x$, then $AE=3\sqrt{2}-x$, $EG=BE-BG=x-\sqrt{2}$,

$\because$ folding triangle $ADE$ along line $DE$, point A lands on the side $BC$, denoted as point F,

$\therefore EA=EF=3\sqrt{2}-x$,

In $\triangle EFG$, $EF^2=EG^2+FG^2$,

i.e., $\left(3\sqrt{2}-x\right)^2=\left(x-\sqrt{2}\right)^2+\left(\sqrt{2}\right)^2$,

solving for $x$, we get $x=\boxed{\frac{7}{4}\sqrt{2}}$


Question: As shown in the figure, the cross section of the dam is a trapezoid with the top width of the dam $DC = 10m$, the height of the dam $15m$, the slope of the incline $AD$ having a gradient ${l}_{1} = 1 : 2$, and the slope of the incline $BC$ having a gradient ${l}_{2} = 3 : 4$. What is the width $AB$ at the bottom of the slope in meters?

Solution:
\textbf{Solution:} As shown in the diagram, draw $DE \perp AB$ at point E through point D, and draw $CF \perp AB$ at point F through point C,\\
then $DE = CF = 15\,m$, and quadrilateral $DEFC$ is a rectangle,\\
$\therefore EF = DC = 10\,m$,\\
$\because$ the slope of ramp $AD$ has a gradient $l_{1} = 1\colon 2$, and the slope of ramp $BC$ has a gradient $l_{2} = 3\colon 4$,\\
$\therefore \frac{DE}{AE} = \frac{1}{2}$, $\frac{CF}{BF} = \frac{3}{4}$, that is, $\frac{15}{AE} = \frac{1}{2}$, $\frac{15}{BF} = \frac{3}{4}$,\\
solving these, we get $AE = 30\,m$, $BF = 20\,m$,\\
therefore, the width at the bottom of the slope $AB = AE + EF + BF = 30 + 10 + 20 = \boxed{60}\,m$.


Question: As shown in the figure, it is known that diamond $ABCD$ has a side length of 2, and $\angle BAD=60^\circ$. If $DE\perp AB$ with the foot of the perpendicular at point $E$, then what is the length of $DE$?

Solution:
\textbf{Solution:} Since $DE \perp AB$,\\
hence $\angle AED = 90^\circ$,\\
in $\triangle ADE$, given $\angle BAD = 60^\circ$, and $AD = 2$,\\
hence $\sin60^\circ = \frac{DE}{AD}$,\\
thus $DE = AD \cdot \sin60^\circ = 2 \times \frac{\sqrt{3}}{2} = \boxed{\sqrt{3}}$.


Question: To calculate $\tan 15^\circ$, as shown in the figure, in $\text{Rt}\triangle ABC$, $\angle C=90^\circ$, $\angle ABC=30^\circ$. Extend $CB$ to $D$, so that $BD=AB$, and connect $AD$, then $\angle D=15^\circ$. Thus, $\tan 15^\circ =\frac{AC}{CD}=\frac{1}{2+\sqrt{3}}=\frac{2-\sqrt{3}}{(2+\sqrt{3})(2-\sqrt{3})}=2-\sqrt{3}$. Following this method, what is the value of $\tan 22.5^\circ$?

Solution:
\textbf{Solution:} As shown in the figure, in the isosceles right-angled $\triangle ABC$, $\angle C=90^\circ$. Extend $CB$ to point $D$ such that $AB=BD$, then $\angle BAD=\angle D$.\\
$\because \angle ABC=45^\circ$,\\
$\therefore 45^\circ=\angle BAD+\angle D=2\angle D$,\\
$\therefore \angle D=22.5^\circ$,\\
Let $AC=1$, then $BC=1$, $AB=\sqrt{2}AC=\sqrt{2}$,\\
$\therefore CD=CB+BD=CB+AB=1+\sqrt{2}$,\\
$\therefore \tan 22.5^\circ=\tan D=\frac{AC}{CD}=\frac{1}{1+\sqrt{2}}=\frac{1-\sqrt{2}}{(1+\sqrt{2})(1-\sqrt{2})}=\boxed{\sqrt{2}-1}$.


Question: As shown in the figure, isosceles right triangle $ABC$ has $\angle C=90^\circ$, with $AC=BC=4$. $M$ is the midpoint of $AB$, and $\angle PMQ=45^\circ$, where the two sides of $\angle PMQ$ intersect $BC$ at point $P$ and $AC$ at point $Q$ respectively. If $BP=3$, what is the length of $AQ$?

Solution:
\textbf{Solution:} As shown in the diagram, connect CM, draw $PF\perp CM$ at point F through point P, and draw $MD\perp AC$ at point D through point M.

In $\triangle ABC$, $AB=\sqrt{AC^2+BC^2}=\sqrt{4^2+4^2}=4\sqrt{2}$,

Since M is the midpoint of AB,

Then $BM=AM=\frac{1}{2}AB=2\sqrt{2}$,

Since $AC=BC$,

Then $CM\perp AB$, $\angle 1=\angle 2=\frac{1}{2}\angle ACB=45^\circ$,

Since in $\triangle BMC$, $\tan\angle 1=\frac{BM}{CM}=1$,

Then $CM=BM=2\sqrt{2}$,

Since $BP=3$,

Then $PC=BC-BP=4-3=1$,

In $\triangle PFC$, $\sin\angle 1=\frac{PF}{PC}=\frac{\sqrt{2}}{2}$, $\cos\angle 1=\frac{CF}{PC}=\frac{\sqrt{2}}{2}$,

Therefore, $PF=\frac{\sqrt{2}}{2}$, $CF=\frac{\sqrt{2}}{2}$,

Therefore, $FM=\frac{3\sqrt{2}}{2}$,

In $\triangle CMD$, $\sin\angle 2=\frac{DM}{CM}=\frac{\sqrt{2}}{2}$, $\cos\angle 2=\frac{CD}{CM}=\frac{\sqrt{2}}{2}$,

Therefore, $DM=2$, $CD=2$,

Therefore, $AD=AC-CD=4-2=2$,

Since $\angle PMQ=45^\circ$,

Then $\angle 3+\angle 4=45^\circ$,

Since $\angle CMD=90^\circ-\angle 2=45^\circ$,

Then $\angle 4+\angle 5=45^\circ$,

Therefore, $\angle 3=\angle 5$,

Since in $\triangle MPF$, $\tan\angle 3=\frac{PF}{FM}=\frac{\frac{\sqrt{2}}{2}}{\frac{3\sqrt{2}}{2}}=\frac{1}{3}$,

Then in $\triangle MDQ$, $\tan\angle 5=\frac{DQ}{MD}=\frac{1}{3}$,

Therefore, $DQ=\frac{1}{3}\times MD=\frac{1}{3}\times 2=\frac{2}{3}$,

Therefore, $AQ=AD+DQ=2+\frac{2}{3}=\boxed{2\frac{2}{3}}$.


Question: As shown in the figure, in $ \triangle ABC$, $ \angle ACB=90^\circ$, $AC=BC$, points D and E are on AB and AC respectively. CD and ED are connected, $ED=CD$, $\tan\angle ADE=\frac{1}{3}$, $BD=\sqrt{2}$. What is the length of $AB$?

Solution:
\textbf{Solution:} As shown in the figure, draw $EH \perp AB$ at H, and draw $DM \perp AC$ at M, and draw $DG \perp BC$ at G, \\
$\because \angle ACB=90^\circ$ and $AC=BC$, \\
$\therefore$ quadrilateral $CGDM$ is a rectangle, $\angle A=\angle B=45^\circ$, \\
$\therefore DG=CM$, \\
$\because BD=\sqrt{2}$, \\
$\therefore DG=BG=BD\cdot \sin 45^\circ=\sqrt{2}\times \frac{\sqrt{2}}{2}=1$, \\
$\therefore CM=1$, \\
$\because DE=DC$ and $DM\perp AC$, \\
$\therefore EM=CM=1$, \\
$\because$ $\tan \angle ADE=\frac{1}{3}=\frac{EH}{DH}$, \\
Let $EH=x$, then $DH=3x$, \\
$\because \angle A=45^\circ$ and $EH\perp AH$, \\
$\therefore AH=EH=x$, \\
From $\sin A=\sin 45^\circ=\frac{AH}{AE}=\frac{\sqrt{2}}{2}$, \\
$\therefore AE=\sqrt{2}x$, \\
Similarly: $\cos A=\cos 45^\circ=\frac{AC}{AB}=\frac{\sqrt{2}}{2}$, \\
$\therefore AB=\sqrt{2}AC$, \\
$\therefore x+3x+\sqrt{2}=\sqrt{2}\left(\sqrt{2}x+2\right)$, \\
Solving gives: $x=\frac{\sqrt{2}}{2}$, \\
$\therefore AB=4x+\sqrt{2}=4\times \frac{\sqrt{2}}{2}+\sqrt{2}=\boxed{3\sqrt{2}}$


Question: As shown in the figure, in the right triangle $\triangle ABC$, the height $AD=4$ on the hypotenuse $BC$, and $\cos B =\frac{4}{5}$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given in right triangle $ABC$, $AD \perp BC$,\\
$\therefore \angle BAD + \angle B = \angle BAD + \angle DAC = 90^\circ$,\\
$\therefore \angle B = \angle DAC$,\\
Since $\cos B = \frac{4}{5}$,\\
$\therefore \cos \angle DAC = \frac{AD}{AC} = \frac{4}{5}$,\\
Given that $AD = 4$,\\
$\therefore AC = \boxed{5}$.


Question: As shown in the diagram, in rectangle $ABCD$, we have $AB=4$ and $AE=\frac{1}{3}AD$. After folding $\triangle ABE$ along $BE$, $\triangle GBE$ is obtained. Extending $BG$ to intersect $CD$ at point $F$, if $F$ is the midpoint of $CD$, what is the length of $BC$?

Solution:
\textbf{Solution:} Extend BF to meet the extension of AD at point H, \\
$\because$ Quadrilateral ABCD is a rectangle, \\
$\therefore$ AD=BC, AD$\parallel$BC, $\angle$A=$\angle$BCF=90$^\circ$, \\
$\therefore$ $\angle$H=$\angle$CBF, \\
In $\triangle BCF$ and $\triangle HDF$, \\
$ \left\{\begin{array}{l}\angle CBF=\angle H \\ \angle BCF=\angle DFH \\ CF=DF \end{array}\right.$  , \\
$\therefore$ $\triangle BCF \cong \triangle HDF$ (AAS), \\
$\therefore$ BC=DH, \\
$\because$ Folding $\triangle ABE$ along BE yields $\triangle GBE$, \\
$\therefore$ $\angle$A=$\angle$BGE=90$^\circ$, AE=EG, \\
$\therefore$ $\angle$EGH=90$^\circ$, \\
$\because$ AE=$ \frac{1}{3}$AD, \\
$\therefore$ Let AE=EG=x, then AD=BC=DH=3x, \\
$\therefore$ ED=2x, \\
$\therefore$ EH=ED+DH=5x, \\
In right $\triangle EGH$, $\sin\angle H=\frac{EG}{EH}=\frac{x}{5x}=\frac{1}{5}$  , \\
$\therefore$ $\sin\angle$CBF=$ \frac{CF}{BF}=\frac{1}{5}$  , \\
$\because$ AB=CD=4, and F is the midpoint of CD, \\
$\therefore$ CF=2, \\
$\therefore$ $\frac{2}{BF}=\frac{1}{5}$ , \\
$\therefore$ BF=10, upon verification, this is consistent with the problem statement, \\
$\therefore$ BC=$ \sqrt{BF^{2}-CF^{2}}=\sqrt{10^{2}-{2}^{2}}=\boxed{4\sqrt{6}}$,


Question: In figure, in $\triangle ABC$, $AD$ is the height from $A$ to side $BC$, $\cos C = \frac{1}{2}$, $AB=10$, and $AC=6$, then what is the length of $BC$?

Solution:
\textbf{Solution:} In right triangle $\triangle ADC$, we have $CD = \cos C \times AC = \frac{1}{2}AC = \frac{1}{2} \times 6 = 3$,\\
therefore $AD = \sqrt{AC^{2} - CD^{2}} = \sqrt{6^{2} - 3^{2}} = 3\sqrt{3}$,\\
In right triangle $\triangle ADB$, $BD = \sqrt{AB^{2} - AD^{2}} = \sqrt{10^{2} - (3\sqrt{3})^{2}} = \sqrt{73}$,\\
therefore $BC = CD + BD = 3 + \sqrt{73}$,\\
Thus, the answer is: $BC = \boxed{3 + \sqrt{73}}$.


Question: As shown in the figure, AB is the diameter of semicircle O, and point C is a point on the semicircle. CD$\perp$AB at point D. If AB=10 and CD=4, what is the value of $\sin\angle BCD$?

Solution:
\textbf{Solution:} As shown in the figure, connect OC, \\
since AB is the diameter of the semicircle O with AB=10, \\
thus OC=OB=5, \\
since CD$\perp$AB at point D, and CD=4, \\
thus OD=$\sqrt{OC^{2}-CD^{2}}=\sqrt{5^{2}-4^{2}}=3$, \\
thus BD=OB−OD=5−3=2, \\
thus BC=$\sqrt{CD^{2}+BD^{2}}=\sqrt{4^{2}+2^{2}}=2\sqrt{5}$, \\
thus $\sin\angle BCD=\frac{BD}{BC}=\frac{2}{2\sqrt{5}}=\boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the diagram, it is known that the radius of $\odot O$ is 2, points A and B are on $\odot O$, $\angle BAC=90^\circ$, and $\tan\angle ACB=2$. What is the maximum value of line segment OC?

Solution:
\textbf{Solution:} As shown in the figure, connect OA, OB, and draw AD$\perp$AO such that $\angle ADO=\angle ABC$,\\
$\because$ in $\triangle ABC$, $\angle BAC=90^\circ$, $\tan \angle ACB=2$,\\
$\therefore$ $\frac{AB}{AC}=2$, i.e., $AB=2AC$,\\
$\because$ $\angle ADO=\angle ABC$, $\angle OAD=\angle CAB=90^\circ$,\\
$\therefore$ $\triangle ADO\sim \triangle ABC$,\\
$\therefore$ $\frac{AO}{AC}=\frac{AD}{AB}$, $\angle AOD=\angle ACB$,\\
$\because$ $\angle DAO=\angle BAC$,\\
$\therefore$ $\angle DAB=\angle OAC$,\\
$\therefore$ $\triangle DAB\sim \triangle OAC$,\\
$\therefore$ $\frac{BD}{OC}=\frac{AB}{AC}=2$,\\
$\therefore$ $OC=\frac{1}{2}BD$,\\
In $\triangle AOD$, $\tan \angle AOD=\tan \angle ACB=\frac{AD}{OA}=2$,\\
$\therefore$ $AD=2AO=4$,\\
$\therefore$ $OD=\sqrt{OA^2+AD^2}=2\sqrt{5}$,\\
$\because$ $BD\le OD+OB=2+2\sqrt{5}$,\\
$\therefore$ the maximum value of BD is $2+2\sqrt{5}$,\\
$\therefore$ the maximum value of OC is $\boxed{1+\sqrt{5}}$.


Question: As shown in the figure, there are two squares with side length of 1 forming a $2\times 1$ lattice graph. Points $A$, $B$, $C$, $D$ are all on the lattice points. Lines $AB$ and $CD$ intersect at point $P$. What is the value of $\tan\angle BPD$?

Solution:
\textbf{Solution:} As shown in the figure, draw $DH \perp BP$ at point $H$,\\
since the side length of the square is 1,\\
therefore $AD=2$,\\
in $\triangle BCD$, $CD=\sqrt{1^2+1^2}=\sqrt{2}$,\\
in $\triangle ABD$, $AB=\sqrt{2^2+1^2}=\sqrt{5}$,\\
since $\frac{1}{2}AB\cdot DH=\frac{1}{2}AD\cdot BD$,\\
therefore $DH=\frac{2\sqrt{5}}{5}$,\\
since BC$\parallel$AD,\\
therefore $\angle DAB=\angle CBP$, $\angle ADP=\angle BCP$,\\
therefore, $\triangle APD\sim \triangle BPC$,\\
therefore $\frac{DP}{CP}=\frac{AD}{BC}=\frac{2}{1}$,\\
therefore DP=2PC,\\
therefore $PD=\frac{2}{3}CD=\frac{2\sqrt{2}}{3}$,\\
in $\triangle PHD$,\\
$PH=\sqrt{PD^2-DH^2}=\frac{2\sqrt{5}}{15}$,\\
therefore $\tan\angle BPD=\frac{DH}{PH}=\boxed{3}$.


Question: As shown in the figure, the regular hexagon $ABCDEF$ is inscribed in circle $\odot O$ with a radius of 2 cm. How many centimeters is $CD$? $G$ is the midpoint of $CD$. Connect $AG$. How many centimeters is $AG$?

Solution:
\textbf{Solution:} Let us connect $AD$, $AC$, $OC$, as depicted:\\
Since the regular hexagon $ABCDEF$ is inscribed in $\odot O$,\\
it follows that $AD$ is the diameter of $\odot O$, and $\angle COD=60^\circ$,\\
which implies $\angle ACD=90^\circ$,\\
Given that $OC=OD=2\text{cm}$,\\
it follows that $\triangle COD$ is an equilateral triangle,\\
thus, $CD=OC=2\text{cm}$, and $\angle ADC=60^\circ$,\\
therefore, $AC=CD\cdot \tan\angle ADC=2\sqrt{3}\text{cm}$,\\
Since $G$ is the midpoint of $CD$,\\
we have $CG=\frac{1}{2}CD=1\text{cm}$,\\
therefore, $AG=\sqrt{AC^2+CG^2}=\boxed{\sqrt{13}}\text{cm}$;


Question: As illustrated, points A, B, C are located at the intersections of the grid lines of a square. If $\triangle ABC$ is rotated counterclockwise around point A to obtain $\triangle AC'B'$, then what is the value of $\tan B''$?

Solution: \textbf{Solution:} To solve, rotate $\triangle ABC$ counter-clockwise about point A to get $\triangle AC'B'$. Based on the characteristic of rotation, $\angle B=\angle B'$. In $\triangle ABC$, points A, B, and C are at the intersections of the square grid lines, thus $\tan B' = \tan B= \boxed{\frac{1}{3}}$.


Question: As shown in the figure, rectangle $ABCD$ is folded along $CE$ such that point B falls exactly on point F on $AD$. If $AB\colon BC=2\colon 3$, then what is the value of $\cos\angle DCF$?

Solution:
\textbf{Solution:} Since rectangle $ABCD$ has $AB:BC=2:3$,\\
let $AB=2m$, then $BC=3m$,\\
thus $AB=CD=2m$, and $AD=BC=CF=3m$, $\angle D=90^\circ$,\\
therefore $\cos\angle DCF=\frac{DC}{FC}=\frac{2m}{3m}=\boxed{\frac{2}{3}}$.


Question: As shown in the figure, within a grid composed of small squares with side length 1, there is a circle $\odot O$ with radius 1 centered on a grid point. What is the value of the tangent of $\angle AED$?

Solution: \textbf{Solution:} According to the inscribed angle theorem, we have $\angle AED = \angle ABC$, thus $\tan\angle AED = \tan\angle ABC = \frac{AC}{AB} = \boxed{\frac{1}{2}}$.


Question: As shown in the figure, within a $5\times4$ square grid, where each small square has a side length of $1$, the vertices of $\triangle ABC$ are all located at the vertices of these small squares. What is the value of $\tan\angle ABC$?

Solution:
\textbf{Solution:} Let point $D$ be chosen on the grid, such that $\angle BDC = 90^\circ$,\\
since $CD = 4$, and $BD = 1$\\
therefore, $\tan\angle ABC = \frac{CD}{BD} = \frac{4}{1} = \boxed{4}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are all on the grid points of the graph paper. What is the value of $\sin C$?

Solution:
Solution: Let the length of each small square in the grid be 1,\\
thus CD=3, and AD=4,\\
$\therefore$ in $\mathrm{Rt}\triangle ACD$, we have $AC=\sqrt{AD^{2}+CD^{2}}=\sqrt{4^{2}+3^{2}}=5$\\
In $\mathrm{Rt}\triangle BCD$, $CB=\sqrt{BD^{2}+CD^{2}}=\sqrt{1^{2}+3^{2}}=\sqrt{10}$\\
As shown in the figure below, draw a perpendicular from point C to AB, and let it intersect the extension of AB at point D. Let AF=4, connect BF, which intersects AC at E,\\
In $\triangle FAB$ and $\triangle ADC$,\\
$\because$ AF=DA=4, $\angle FAB=\angle ADC=90^\circ$, AB=DC=3,\\
$\therefore \triangle FAB \cong \triangle ADC$ (SAS),\\
$\therefore \angle F=\angle CAD$, FB=AC=5\\
$\because \angle FBA=\angle ABE$,\\
$\therefore \triangle ABE \sim \triangle FBA$,\\
$\therefore \angle AEB=\angle FAB=90^\circ$,\\
$\therefore \frac{AB}{EB}=\frac{FB}{AB}$, that is $\frac{3}{EB}=\frac{5}{3}$\\
$\therefore EB=\frac{9}{5}$\\
$\therefore \sin C= \frac{BE}{BC}=\frac{\frac{9}{5}}{\sqrt{10}}=\boxed{\frac{9\sqrt{10}}{50}}$


Question: As shown in the figure, in the Cartesian coordinate system, there is a right-angled triangle $Rt\triangle AOB$, with $\angle ABO=90^\circ$ and $\angle AOB=30^\circ$. The right-angle side $OB$ is on the positive half of the $y$-axis, point $A$ is in the first quadrant, and $OA=1$. Rotate the $Rt\triangle AOB$ counter-clockwise around the origin by $30^\circ$, and at the same time, double the lengths of its sides (i.e., $OA_{1}=2OA$). This results in $Rt\triangle OA_{1}B_{1}$. Similarly, rotate $Rt\triangle OA_{1}B_{1}$ around the origin $O$ counter-clockwise by $30^\circ$, and at the same time, double the lengths of its sides, resulting in $Rt\triangle OA_{2}B_{2}$, ..., Following this rule, we obtain $Rt\triangle OA_{2021}B_{2021}$. What is the length of $OB_{2021}$?

Solution:
\textbf{Solution:} In right triangle $\triangle AOB$, we have $\angle AOB=30^\circ$ and $OA = 1$, \\
therefore, $OB = OA \cdot \cos{\angle AOB} = \frac{\sqrt{3}}{2}$, \\
according to the problem, $OB_1 = 2OB = \frac{\sqrt{3}}{2} \times 2$, \\
$OB_2 = 2OB_1 = \frac{\sqrt{3}}{2} \times 2^2$, \ldots \\
$OB_n = \frac{\sqrt{3}}{2} \times 2^n = \sqrt{3} \times 2^{n-1}$, \\
therefore, the length of $OB_{2021}$ is: $\sqrt{3} \times 2^{2020} = \boxed{\sqrt{3} \times 2^{2020}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB = 90^\circ$, $\angle A = 30^\circ$, $AC = 15\, \mathrm{cm}$, point $O$ is on the median $CD$, when a circle with radius $3\, \mathrm{cm}$ centered at $O$ is tangent to the sides of $\triangle ABC$, what is the length of $OC$?

Solution:
\textbf{Solution:} In $\triangle ABC$, $\angle ACB=90^\circ$, $\angle A=30^\circ$,\\
$\therefore \angle B=60^\circ$,\\
$\because AC = 15\, \text{cm}$,\\
$\therefore \cos A=\frac{AC}{AB}$,\\
$\therefore AB=\frac{AC}{\cos A}=\frac{15}{\frac{\sqrt{3}}{2}}=10\sqrt{3}$,\\
$\because CD$ is the median to side $AB$,\\
$\therefore CD=AD=BD=\frac{1}{2}AB=5\sqrt{3}$,\\
$\therefore \angle BDC=\angle BCD=\angle B=60^\circ$, $\angle ACD=\angle A=30^\circ$,\\
(1) When circle $O$ is tangent to $AB$, draw $OE\perp AB$ at $E$,\\
In $\triangle ODE$, $\angle BDC=60^\circ$, $OE=3$,\\
$\therefore \sin\angle BDC=\frac{OE}{OD}$,\\
$\therefore OD=\frac{OE}{\sin\angle BDC}=\frac{3}{\frac{\sqrt{3}}{2}}=2\sqrt{3}$;\\
$\therefore OC=CD-OD=5\sqrt{3}-2\sqrt{3}=3\sqrt{3}$;\\
(2) When circle $O$ is tangent to $BC$, draw $OE\perp BC$,\\
In $\triangle OCE$, $\angle BCD=60^\circ$, $OE=3$,\\
$\therefore \sin\angle BCD=\frac{OE}{OC}$,\\
$\therefore OC=\frac{OE}{\sin\angle BCD}=\frac{3}{\frac{\sqrt{3}}{2}}=2\sqrt{3}$;\\
(3) When circle $O$ is tangent to $AC$, draw $OE\perp AC$ at $E$,\\
In $\triangle OCE$, $\angle ACD=30^\circ$, $OE=3$,\\
$\therefore \sin\angle ACD=\frac{OE}{OC}$,\\
$\therefore OC=\frac{OE}{\sin\angle ACD}=\frac{3}{\frac{1}{2}}=\boxed{6}$.


Question: As shown in the figure, in $\triangle ABC$, $I$ is the incenter of $\triangle ABC$, $O$ is a point on side $AB$, and $\odot O$ passes through point $B$ and is tangent to $AI$ at point $I$. If $\tan\angle BAC = \frac{24}{7}$, then what is the value of $\sin\angle ACB$?

Solution:
\textbf{Solution:} As shown in the figure, extend line OI to intersect AC at D and connect BI.\\
$\because$ AI is tangent to the circle with center O,\\
$\therefore$ AI $\perp$ OD,\\
$\therefore$ $\angle$AIO = $\angle$AID = $90^\circ$,\\
$\because$ I is the incenter of $\triangle ABC$,\\
$\therefore$ $\angle$OAI = $\angle$DAI, $\angle$ABI = $\angle$CBI,\\
$\because$ AI = AI,\\
$\therefore$ $\triangle AOI \cong \triangle ADI$ (ASA),\\
$\therefore$ AO = AD,\\
$\because$ OB = OI,\\
$\therefore$ $\angle$OBI = $\angle$OIB,\\
$\therefore$ $\angle$OIB = $\angle$CBI,\\
$\therefore$ OD $\parallel$ BC,\\
$\therefore$ $\angle$ADO = $\angle$C,\\
Construct OE $\perp$ AC at E,\\
$\because$ $\tan\angle$BAC = $\frac{OE}{AE}$ = $\frac{24}{7}$,\\
$\therefore$ assume OE = 24k, AE = 7k without loss of generality,\\
$\therefore$ OA = AD = 25k,\\
$\therefore$ DE = AD - AE = 18k,\\
$\therefore$ OD = $\sqrt{OE^2 + DE^2}$ = 30k,\\
$\therefore$ $\sin\angle$ACB = $\frac{OE}{OD}$ = $\frac{24k}{30k}$ = $\boxed{\frac{4}{5}}$.


Question: As shown in the diagram, in diamond $ABCD$, $AE \perp BC$ with $E$ being the foot of the perpendicular. If $\cos B = \frac{4}{5}$ and $EC = 2$, and $P$ is a moving point on side $AB$, then what is the minimum length of segment $PE$?

Solution:
\textbf{Solution:} Let the side length of the rhombus be $a$, \\
$\therefore BE = a - 2$, \\
$\because AE \perp BC$, \\
$\therefore \angle AEB = 90^\circ$, \\
$\because \cos B = \frac{4}{5}$ $\therefore \sin B = \frac{3}{5}$, \\
$\cos B = \frac{BE}{AB} = \frac{a - 2}{a} = \frac{4}{5}$, \\
$\therefore a = 10$, \\
$\therefore BE = 8$, \\
$\because$ when $EP$ is perpendicular to $AB$, the length is at its minimum, \\
$\therefore$ in $\triangle BPE$, $\sin B = \frac{PE}{BE} = \frac{3}{5}$, \\
$\therefore PE = \frac{3}{5} \times 8 = \boxed{4.8}$, \\
$\therefore$ the minimum length of segment $PE$ is $4.8$.


Question: As shown in the figure, circle $ \odot O$ intersects with side $ OA$ and $ OE$ of the regular hexagon $OABCDE$ at points $F$ and $G$, respectively. Point $M$ is the midpoint of the minor arc $FG$. If $FM=4\sqrt{2}$, what is the distance from point $O$ to line $FM$?

Solution:
\textbf{Solution:} As shown in the figure, connect OM, and draw OH$\perp$FM through point O,\\
$\because$ hexagon ABCDEO is a regular hexagon,\\
$\therefore$ $\angle$AOE=120$^\circ$,\\
$\because$ point M is the midpoint of the minor arc FG,\\
$\therefore$ $\angle$AOM=60$^\circ$,\\
$\because$ OM=OF,\\
$\therefore$ $\triangle$OMF is an equilateral triangle,\\
$\therefore$ MH=FH=$\frac{1}{2}$MF=2$\sqrt{2}$, $\angle$OFH=60$^\circ$,\\
$\therefore$ OH=$\tan60^\circ\cdot$ HF=$\sqrt{3}\times 2\sqrt{2}=\boxed{2\sqrt{6}}$.


Question: As shown in the figure, in right triangle \( \triangle ABC \), where \( \angle ACB=90^\circ \), \( CD \) is the median to the hypotenuse \( AB \), and it is given that \( CD=2 \) and \( AC=3 \), what is the value of \( \cos A \)?

Solution:
\textbf{Solution:} Since $D$ is the midpoint of $AB$, and $CD=2$, \\
$\therefore$ $AC=2CD=4$, \\
$\cos A= \frac{AC}{AB}=\boxed{\frac{3}{4}}$.


Question: As shown in the figure, the detector on the hot air balloon indicates that the angle of depression from the bottom of the hot air balloon at point $A$ to the top of a building is $30^\circ$, and the angle of depression to the bottom of the building is $60^\circ$. The distance from point $A$ on the hot air balloon to the ground is $150m$. What is the height of the building in meters?

Solution:
\textbf{Solution:} As shown in the diagram, draw $AH\perp BC$, extending through point $CB$ to $H$,\\
In right $\triangle ACD$,\\
$\because \angle CAD=30^\circ$, $AD=150m$\\
$\therefore CD=AD\cdot \tan 30^\circ=150\times \frac{\sqrt{3}}{3}=50\sqrt{3}(m)$,\\
$\therefore AH=CD=50\sqrt{3}m$\\
In right $\triangle ABH$,\\
$\because \angle BAH=30^\circ$, $AE=50\sqrt{3}m$\\
$\therefore BH=AH\cdot \tan 30^\circ=50\sqrt{3}\times \frac{\sqrt{3}}{3}=50(m)$,\\
$\therefore BC=AD−BH=150−50=\boxed{100}(m)$


Question: As shown in the figure, point C is any point on the semicircle with AB as the diameter, where $AB=8$. After connecting AC, the line segment AC is rotated $120^\circ$ counterclockwise around point A to obtain the line segment $AC'$. What is the maximum value of $BC'$?

Solution:
\textbf{Solution:} As shown in the figure, find the midpoint $O$ of the line segment $AB$, connect $OC$, and rotating the line segment $AC$ around point $A$ by $120^\circ$ can be considered as rotating $\triangle AOC$ around point $A$ by $120^\circ$ to obtain $\triangle AO'C'$,\\
$\therefore \triangle AOC \cong \triangle AO'C'$,\\
$\therefore OC=O'C'=4$,\\
Taking point $O'$ as the center and $O'C'$ as the radius to draw a circle, when points $O'$, $C'$, and $B$ are collinear, $BC'$ takes the maximum value,\\
As shown in the figure: $\angle DAO' =60^\circ$, $O'A=4$, $AB=8$,\\
$AD=O'A\cdot \cos 60^\circ=2$, $O'D=2\sqrt{3}$,\\
$\therefore BO'=\sqrt{DB^2+O'B^2}=\sqrt{10^2+(2\sqrt{3})^2}=4\sqrt{7}$,\\
$\therefore BC' =BO'+C'O' =4\sqrt{7}+4$.\\
Therefore, the answer is: $BC'=\boxed{4\sqrt{7}+4}$.


Question: As shown in the diagram, by connecting two non-adjacent vertices within a regular hexagon with side length $a$, a shape with a hexagonal interior is formed, as illustrated. What is the perimeter of the shaded area?

Solution:
\textbf{Solution:} Let $PH \perp AB$ at H,\\
since the outer hexagon is a regular hexagon with side length $a$,\\
therefore, the perimeter of the outer hexagon is $6a$,\\
since the hexagon is regular,\\
therefore $\angle CAB=\frac{(6-2)\times 180^\circ}{6}=120^\circ$, CA=BA,\\
therefore $\angle ABP=30^\circ$,\\
also, since PA=PB,\\
therefore $\angle PAB=30^\circ$,\\
therefore $PA=\frac{AH}{\cos\angle PAH}=\frac{\sqrt{3}}{3}a$,\\
as the problem states, the shaded area is a regular hexagon,\\
therefore, the perimeter of the shaded area is $\frac{\sqrt{3}}{3}a\times 6=2\sqrt{3}a$.\\
Hence, the answer is: $2\sqrt{3}a=\boxed{2\sqrt{3}a}$.


Question: As shown in the figure, if the vertices of $\triangle ABC$ are lattice points on a square grid, what is the value of $\cos\angle C$?

Solution:
\textbf{Solution:} As shown in the figure, connect BE in the diagram, \\
by the Pythagorean theorem, we find: $CE=\sqrt{4^2+2^2}=2\sqrt{5}$, $BE=\sqrt{1^2+2^2}=\sqrt{5}$, $BC=\sqrt{3^2+4^2}=5$, \\
$\therefore CE^2+BE^2=(2\sqrt{5})^2+(\sqrt{5})^2=25=BC^2$, \\
$\therefore \triangle CEB$ is a right triangle, $\angle CEB=90^\circ$, \\
$\therefore \cos C=\frac{CE}{CB}=\frac{2\sqrt{5}}{5}=\boxed{\frac{2\sqrt{5}}{5}}$.


Question: As shown in Figure 1, a moving point $P$ starts from vertex $A$ of the rhombus $ABCD$, moves along the path $A \to C \to D$ at a speed of $1cm/s$, and stops at point $D$. Let the movement time of point $P$ be $x(s)$, and the area of $\triangle PAB$ be $y(cm^2)$. The graph showing the functional relationship between $y$ and $x$ is shown in Figure 2. What is the value of $a$?

Solution:
\textbf{Solution:} Given the graph of the function, we can conclude that:\\
When $x=4s$, points P and C coincide, and area $S\triangle ABP=a$, which is the maximum area,\\
$\therefore AC=1\times 4=4$,\\
When $x=a+4$, points P and D coincide,\\
$\therefore AB=CD=1\times (a+4)-4=a$,\\
As shown in the diagram, draw CK$\perp$AB at K through C,\\
$\therefore \frac{1}{2}a\cdot CK=a$,\\
$\therefore CK=2$,\\
$\therefore \sin\angle CAK=\frac{CK}{CA}=\frac{1}{2}$,\\
$\therefore \angle CAK=30^\circ$,\\
$\therefore \angle ACK=60^\circ$,\\
Given that rhombus ABCD,\\
$\therefore AB=BC=a$, $\angle BCA=\angle BAC=30^\circ$,\\
$\therefore \angle BCK=60^\circ-30^\circ=30^\circ$,\\
Since $\cos\angle BCK=\frac{CK}{BC}$,\\
$\therefore \frac{2}{a}=\cos30^\circ=\frac{\sqrt{3}}{2}$,\\
$\therefore \sqrt{3}a=4$,\\
$\therefore a=\boxed{\frac{4\sqrt{3}}{3}}$.


Question: In the figure, in the equilateral triangle $ABC$ with a side length of $2\sqrt{3}$, the moving points $D$ and $E$ are on the sides $BC$ and $AC$, respectively, maintaining $AE=CD$. Lines $BE$ and $AD$ are drawn and intersect at point $P$. What is the minimum value of $CP$?

Solution:
\textbf{Solution:} As shown in the diagram:

$\because$ CD = AE, BC = AC,

$\therefore$ BD = CE,

In $\triangle ABD$ and $\triangle BCE$,

$\left\{\begin{array}{l}
AB=BC\\
\angle ABD=\angle BCE\\
BD=CE
\end{array}\right.$,

$\therefore \triangle ABD \cong \triangle BCE$ (SAS),

$\therefore \angle BAD=\angle CBE$,

$\because \angle APE=\angle ABE+\angle BAD$, $\angle APE=\angle BPD$, $\angle ABE+\angle CBE=60^\circ$,

$\therefore \angle BPD=\angle APE=\angle ABC=60^\circ$,

$\therefore \angle APB=120^\circ$,

$\therefore$ the path of point P is $\overset{\frown}{AB}$, with $\angle AOB=120^\circ$, connect CO,

$\because$ OA = OB, CA = CB, OC = OC,

$\therefore \triangle AOC \cong \triangle BOC$ (SSS),

$\therefore \angle OAC=\angle OBC$, $\angle ACO=\angle BCO=30^\circ$,

$\because \angle AOB+\angle ACB=180^\circ$,

$\therefore \angle OAC+\angle OBC=180^\circ$,

$\therefore \angle OAC=\angle OBC=90^\circ$,

$\therefore OC=AC \div \cos 30^\circ=4$, OA= $\frac{1}{2}$ OC=2,

$\therefore OP=2$,

$\because PC \geq OC-OP$,

$\therefore PC \geq 2$,

$\therefore$ The minimum value of PC is $\boxed{2}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AB=10$, $BC=6$, E is a point on side $AC$, $AE=5$, $ED\perp AB$, and the foot of the perpendicular is point D. Then, what is the value of $DE$?

Solution: \textbf{Solution:} Since $\angle C=90^\circ$ and $AB=10$, $BC=6$, we have $\sin A=\frac{BC}{AB}=\frac{6}{10}=\frac{3}{5}$. Given that $ED \perp AB$, it follows that $\sin A=\frac{DE}{AE}=\frac{3}{5}$. Hence, $AE=5$ and so $DE=\boxed{3}$.


Question: As shown in the figure, within rectangle $ABCD$, point $E$ is on side $AB$, and triangles $\triangle BEC$ and $\triangle FEC$ are symmetrical about line $EC$. The symmetric point of point $B$, point $F$, is on side $AD$, with $G$ being the midpoint of $CD$. Connecting $BG$ intersects $CE$ and $CF$ at points $M$ and $N$ respectively. If $BM=BE$ and $MG=2$, what is the length of $BN$? What is the value of $\sin\angle AFE$?

Solution: \textbf{Solution:} Given that $BM = BE$, it follows that $\angle BEM = \angle BME$. Since $AB \parallel CD$, we have $\angle BEM = \angle GCM$. Moreover, given $\angle BME = \angle GMC$, it leads to $\angle GCM = \angle GMC$, thus $MG = GC = 2$. Given $G$ is the midpoint of $CD$, it follows that $CD = AB = 4$. Connecting $BF$ and $FM$, by folding, we obtain $\angle FEM = \angle BEM$ and $BE = EF$, thus $BM = EF$. Since $\angle BEM = \angle BME$, it follows $\angle FEM = \angle BME$, thus $EF \parallel BM$. Consequently, quadrilateral $BEFM$ is a parallelogram. Given $BM = BE$, quadrilateral $BEFM$ is a rhombus. Since $\angle EBC = \angle EFC = 90^\circ$ and $EF \parallel BG$, it follows $\angle BNF = 90^\circ$. Given $BF$ bisects $\angle ABN$, it follows $FA = FN$. Hence, right triangle $ABF$ is congruent to right triangle $NBF$ (HL), leading to $BN = AB = \boxed{4}$. Since $FE = FM$, $FA = FN$, and $\angle A = \angle BNF = 90^\circ$, right triangle $AEF$ is congruent to right triangle $NMF$ (HL), leading to $AE = NM$. Let $AE = NM = x$, then $BE = FM = 4 - x$, and $NG = MG - NM = 2 - x$. Since $FM \parallel GC$, triangles $FMN$ and $CGN$ are similar, leading to $\frac{CG}{FM} = \frac{GN}{NM}$, i.e., $\frac{2}{4-x} = \frac{2-x}{x}$. Solving this equation gives $x = 4 + 2\sqrt{2}$ (discard) or $x = 4 - 2\sqrt{2}$, thus $EF = BE = 4 - x = 2\sqrt{2}$. Hence, $\sin\angle AFE = \frac{AE}{EF} = \frac{4 - 2\sqrt{2}}{2\sqrt{2}} = \boxed{\sqrt{2} - 1}$.


Question: As shown in the figure, there is a grid composed of small squares with equal side lengths. Points $A$, $B$, $P$, and $Q$ are all located on the vertices of the square grid. The line segments $AB$ and $PQ$ intersect at point $E$. What is the value of $\tan\angle AEP$?

Solution:
\textbf{Solution:} As shown in the figure, let the side length of the small square be 1,\\
according to the grid characteristics, $\angle PQF=\angle CBF=45^\circ$,\\
$\therefore PQ\parallel BC$,\\
$\therefore \angle QEB=\angle ABC$,\\
$\because \angle AEP=\angle QEB$,\\
$\therefore \angle AEP=\angle ABC$,\\
$\because AB=\sqrt{1^2+3^2}=\sqrt{10}$, $AC=\sqrt{1^2+1^2}=\sqrt{2}$, $BC=\sqrt{2^2+2^2}=\sqrt{8}=2\sqrt{2}$,\\
$\therefore AC^2+BC^2=AB^2$,\\
$\therefore \triangle ABC$ is a right triangle, and $\angle ACB=90^\circ$,\\
$\therefore \tan\angle ABC= \frac{AC}{BC}=\frac{\sqrt{2}}{2\sqrt{2}}=\frac{1}{2}$,\\
$\therefore \tan\angle AEP=\tan\angle ABC= \boxed{\frac{1}{2}}$.


Question: As shown in the figure, AB is the diameter of $\odot O$, C is a point on $\odot O$, and a tangent to $\odot O$ through point C intersects the extension of AB at point E. If $\angle A=30^\circ$, what is the value of $\sin E$?

Solution:
\textbf{Solution:} Connect OC, \\
$\because$ CE is a tangent of $\odot$O, \\
$\therefore$ OC$\perp$CE, \\
$\because$ $\angle$A=30$^\circ$, and OC=OA, \\
$\therefore$ $\angle$BOC=2$\angle$A=60$^\circ$, \\
$\therefore$ $\angle$E=90$^\circ$−$\angle$BOC=30$^\circ$, \\
$\therefore$ $\sin\angle$E=$\sin30^\circ$= \boxed{\frac{1}{2}}.


Question: As shown in the figure, it is known that the radius of sector $OAB$ is $6$, $C$ is any point on arc $AB$ (not coinciding with $A$ or $B$), $CM\perp OA$ with the foot of the perpendicular being $M$, $CN\perp OB$ with the foot of the perpendicular being $N$, connecting $MN$. If $\angle AOB=45^\circ$, what is the length of $MN$?

Solution:
\textbf{Solution:} Connect OC, extend OA and NC to intersect at D, then OC = 6, \\
$\because$ CM $\perp$ OA, CN $\perp$ OB, \\
$\therefore$ $\angle$DMC = $\angle$DNO = $90^\circ$, \\
$\because$ $\angle$D = $\angle$D, \\
$\therefore$ $\triangle$DMC $\sim$ $\triangle$DNO, \\
$\therefore$ $\frac{DM}{DN}=\frac{DC}{DO}$, which implies $\frac{DM}{DC}=\frac{DN}{DO}$, \\
$\because$ $\angle$D = $\angle$D, \\
$\therefore$ $\triangle$DMN $\sim$ $\triangle$DCO, \\
$\therefore$ $\frac{MN}{CO}=\frac{DN}{DO}$, \\
$\because$ CN $\perp$ OB, $\angle$AOB = $45^\circ$, \\
$\therefore$ sin$\angle$AOB = $\frac{DN}{OD}=\frac{\sqrt{2}}{2}$, \\
$\therefore$ $\frac{MN}{OC}=\frac{\sqrt{2}}{2}$, \\
$\because$ OC = 6, \\
$\therefore$ $\frac{MN}{6}=\frac{\sqrt{2}}{2}$, \\
$\therefore$ MN = $\boxed{3\sqrt{2}}$.


Question: As shown in the figure, in the isosceles triangle $ABC$, $\angle C=90^\circ$, $AC=BC=6$, and point $D$ is a point on $AC$, $\tan\angle DBA=\frac{1}{\sqrt{5}}$. What is the length of $AD$?

Solution:
\textbf{Solution:} Draw line $DE \perp AB$ at point E, as shown in the diagram,\\
$\therefore$ $\angle AED=\angle DEB=90^\circ$,\\
$\because$ $\angle C=90^\circ$, $AC=BC=6$,\\
$\therefore$ $AB=\sqrt{AC^2+BC^2}=\sqrt{2}AC=6\sqrt{2}$, $\angle A=45^\circ$,\\
$\therefore$ $\triangle ADE$ is an isosceles right triangle,\\
$\therefore$ $AE=DE$,\\
In $\triangle ADE$, let $AE=x$, then $DE=x$,\\
$\therefore$ $AD=\sqrt{AE^2+DE^2}=\sqrt{2}AE=\sqrt{2}x$\\
In $\triangle BED$, $\tan\angle DBE=\frac{DE}{BE}=\frac{1}{5}$,\\
$\therefore$ $BE=5x$,\\
$\therefore$ $AB=AE+BE=x+5x=6\sqrt{2}$,\\
$\therefore$ $x=\sqrt{2}$\\
$\therefore$ $AD=\sqrt{2}x=\boxed{2}$


Question: As shown in the figure, in quadrilateral $ABCD$, $AB\parallel CD$, $DE$ bisects $\angle ADC$ and intersects $BC$ at point $E$, $AE\perp DE$, $AB=4$, $AE=3$, and $\sin\angle CDE=\frac{3}{5}$. Then, what is the length of $CD$?

Solution:
\textbf{Solution:} As shown in the figure, extend $AE,DC$ to intersect at point $F$,\\
since $DE$ bisects $\angle ADC$ and $DE\perp AE$,\\
therefore $\angle ADE=\angle FDE$, $AE=EF$,\\
let $\angle ADE=\angle FDE=\alpha$\\
since $DE\perp AE$, $\sin\angle CDE=\frac{3}{5}$, $AB=4$, $AE=3$\\
therefore $\frac{AE}{AD}=\sin\alpha =\frac{3}{5}$\\
therefore $AD=5$\\
therefore $DE=\sqrt{AD^{2}-AE^{2}}=4$\\
since $AB\parallel CD$\\
therefore $\angle F=\angle BAE=90^\circ-\alpha$\\
since $DE\perp AE$\\
therefore $\angle DAE=90^\circ-\alpha$,\\
therefore $\angle F=\angle DAE=\angle EAB$\\
therefore $DF=AD=5$\\
In $\triangle ABE$ and $\triangle FCE$,\\
$\left\{\begin{array}{l}
\angle BAE=\angle F\\
\angle AEB=\angle FEC\\
AE=FE
\end{array}\right.$,\\
therefore $\triangle ABE\cong \triangle FCE$,\\
therefore $CF=AB=4$,\\
therefore $DC=DF-CF=5-4=\boxed{1}$,


Question: As shown in the figure, it is known that in the circle $\odot O$ with diameter $AB=8\sqrt{3}$, radius $OC\perp AB$, point D is the trisection point of the semicircle $\overset{\frown}{AB}$, and point P is a moving point on radius $OC$. When the value of $PB+PD$ is minimized, what is the length of $PO$?

Solution:
\textbf{Solution:} Draw $DO$, $DA$ intersects $OC$ at point P,\\
since $OC\perp AB$ and point O is the midpoint of $AB$,\\
therefore, the point symmetrical to B with respect to $OC$ is point A,\\
therefore, the intersection point P of $DA$ and $OC$ makes the value of $PB+PD$ the smallest,\\
since point D is the trisection point of the semicircle $\overset{\frown}{AB}$,\\
therefore, $\angle DOB=60^\circ$,\\
therefore, $\angle DAB=30^\circ$,\\
since $\angle AOP=90^\circ$, $OA=\frac{1}{2}AB$, $AB=8\sqrt{3}$, $\angle PAO=30^\circ$,\\
therefore, $OA=4\sqrt{3}$,\\
therefore, $OP=OA\cdot\tan30^\circ=4\sqrt{3}\times \frac{\sqrt{3}}{3}=\boxed{4}$.


Question: As shown in the figure, a drone interest group is carrying out activities on the playground. At that moment, the drone is at point D, $20\sqrt{3}$ meters above the ground. The drone measures the depression angle from the controller at point A to be $30^\circ$, and the depression angle from the top of the teaching building at point C to be $45^\circ$. Additionally, through manual measurement, the horizontal distance between the controller A and the teaching building along $BC$ is found to be 80 meters. What is the height of the teaching building $BC$? (Note: Points A, B, C, and D are all in the same plane, reference data $\sqrt{3} \approx 1.7$)

Solution:
\textbf{Solution:} Construct $DE\perp AB$ at point E, and $CF\perp DE$ at point F, \\
according to the problem, $DE=20\sqrt{3}$, $\angle A=30^\circ$, $\angle DCF=45^\circ$, \\
in $Rt\triangle ADE$, $\angle AED=90^\circ$, \\
therefore, $\tan30^\circ=\frac{DE}{AE}=\frac{\sqrt{3}}{3}$, \\
therefore, $AE=\sqrt{3}DE=60$ (meters), \\
since $AB=80$ meters, \\
therefore, $BE=80-60=20$ (meters), \\
since $\angle FEB=\angle CBE=\angle CFE=90^\circ$, \\
since quadrilateral $BCFE$ is a rectangle, \\
therefore, $CF=BE=20$ meters, \\
in $Rt\triangle DCF$, $\angle DFC=90^\circ$, \\
therefore, $\angle CDF=\angle DCF=45^\circ$, \\
therefore, $CF=DF=20$, \\
therefore, $BC=EF=DE-DF=20\sqrt{3}-20\approx 20\times 1.7-20=\boxed{14}$ (meters).


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, where $O$ is the origin, point $B$ is on the positive half-axis of the x-axis, and quadrilateral $OBCA$ is a parallelogram. It is given that $\sin\angle AOB=\frac{4}{5}$. The inverse proportion function $y=\frac{m}{x}$ ($m>0$) passes through point $A$ in the first quadrant and intersects $BC$ at point $F$. If point $F$ is the midpoint of $BC$, and the area of $\triangle AOF$ is 12, then what is the value of $m$?

Solution:
\textbf{Solution:} Draw $AM\perp OB$ at $M$, and $FN\perp OB$ at $N$, where $OA=5k$,\\
since $\sin\angle AOB=\frac{4}{5}$,\\
therefore $AM=4k$, $OM=3k$, $m=12k^{2}$,\\
since quadrilateral OACB is a parallelogram and F is the midpoint of $BC$,\\
therefore $FN=2k$, $ON=6k$,\\
since ${S}_{\triangle AOM}={S}_{\triangle OFN}$,\\
${S}_{\text{trapezium} OAFN}={S}_{\text{trapezium} AMNF}+{S}_{\triangle AOM}={S}_{\triangle AOF}+{S}_{\triangle OFN}$,\\
therefore ${S}_{\text{trapezium} AMNF}={S}_{\triangle AOF}=12$,\\
therefore $\frac{1}{2}(4k+2k)\cdot 3k=12$,\\
therefore $k^{2}=\frac{4}{3}$,\\
therefore $m=12k^{2}=\boxed{16}$.


Question: As shown in the diagram, in the right-angled $\triangle ABC$, $\angle C=90^\circ$, $AC=8$, $BC=6$. The point $M$ starts from point $C$ and moves along the segment $CA$ towards point $A$, connecting $BM$, and $MN \perp BM$ intersects side $AB$ at point $N$. If $CM=2$, what is the length of segment $AN$?

Solution:
\textbf{Solution:} As shown in the figure, draw $NH \perp AC$ at point $H$ through point $N$, \\
$\because \angle C=90^\circ$, $AC=8$, $BC=6$, \\
$\therefore AB=\sqrt{AC^2+BC^2}=10$, \\
$\because \sin A=\frac{NH}{AN}=\frac{BC}{AB}=\frac{3}{5}$, $\tan A=\frac{NH}{AH}=\frac{BC}{AC}=\frac{3}{4}$, \\
$\therefore AN=\frac{5}{3}NH$, $AH=\frac{4}{3}NH$, \\
$\because \angle C=\angle BMN=\angle MNH=90^\circ$, \\
$\therefore \angle BMC+\angle NMH=90^\circ$, \\
$\therefore \angle CBM=90^\circ-\angle BMC=\angle NMH$, \\
$\therefore \tan\angle CBM=\frac{CM}{BC}=\tan\angle NHM=\frac{NH}{MH}$, \\
thus, $\frac{2}{6}=\frac{NH}{8-2-\frac{4}{3}NH}$, \\
Solving gives: $NH=\frac{18}{13}$, \\
$\therefore AN=\frac{5}{3}NH=\frac{5}{3} \times \frac{18}{13}=\boxed{\frac{30}{13}}$.


Question: In a rectangular piece of paper $ABCD$, where $AD=10$ and $AB=4\sqrt{3}$, $M$ and $N$ are the midpoints of $AB$ and $CD$, respectively. Now, fold the paper as shown in the figure so that point $B$ falls on point $F$ on $MN$. What is the length of $EF$?

Solution:
\textbf{Solution:} As shown in the diagram, draw $FH\perp BC$ at point H, obtaining rectangle $BMFH$.\\
Since M is the midpoint of $AB$,\\
it follows that $AM=2\sqrt{3}$, $\angle AMF=90^\circ$,\\
therefore, $\angle AFM=30^\circ$,\\
thus, $MF=AF\cdot \cos 30^\circ=6$.\\
Since quadrilateral $BMFH$ is a rectangle,\\
it follows that $BH=MF=6$.\\
Let $EF=x$, then $EH=6-x$, $FH=2\sqrt{3}$.\\
In $\triangle EFH$, $EF^2=EH^2+FH^2$,\\
which gives $x^2=(6-x)^2+(2\sqrt{3})^2$,\\
solving this yields $x=\boxed{4}$.


Question: As shown in the figure, there is a semicircle $O$ with the diameter $AB$ of $\triangle ABC$, and point $C$ is exactly on the semicircle. Draw $CD\perp AB$ at $D$. Given that $\cos\angle ACD=\frac{3}{5}$ and $BC=2$, what is the length of $AC$?

Solution:
\textbf{Solution:} Given that point C is on the semicircle, and AB is the diameter\\
\(\therefore\angle ACB=90^\circ\)\\
Since \(CD\perp AB\), and \(\cos\angle ACD= \frac{3}{5}\)\\
\(\therefore\sin\angle CAB=\cos\angle ACD= \frac{3}{5}\)\\
\(\therefore\cos\angle CAB= \sqrt{1-\sin^2\angle CAB}=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}\)\\
\(\therefore\tan\angle CAB= \frac{\sin\angle CAB}{\cos\angle CAB}=\frac{3}{4}\)\\
\(\therefore \frac{BC}{AC}=\frac{3}{4}\)\\
\(\therefore AC= \frac{4}{3}BC= \frac{4}{3}\times 2=\boxed{\frac{8}{3}}\)


Question: As shown in the figure, points $A$, $P$, and $T$ lie on the circle $\odot O$, $OT\perp TB$, the arc $\overset{\frown}{AT}$ measures $60^\circ$, the radius of $\odot O$ is 2, $TB=2$, point $B$ and point $A$ are on opposite sides of line $OT$, $PB$ intersects $\odot O$ at point $C$. When $\angle APB=60^\circ$, what is the value of $\angle PBT$? What is the length of $BC$?

Solution:
\textbf{Solution}: Connect $OA$, $OC$, $TC$, and draw $CD\perp TB$ at point $C$ with foot $D$,\\
$\because \angle APB=60^\circ$,\\
$\therefore \angle AOC=2\angle APC=120^\circ$,\\
$\because$ the degree of $\overset{\frown}{AT}$ is $60^\circ$,\\
$\therefore \angle AOT=60^\circ$,\\
$\therefore \angle TOC=\angle AOC-\angle AOT=60^\circ$,\\
$\because OT=OC$,\\
$\therefore \triangle OTC$ is an equilateral triangle,\\
$\therefore TC=OT=OC=2$, and $\angle OTC=60^\circ$,\\
$\because OT\perp TB$,\\
$\therefore \angle OTB=90^\circ$,\\
$\therefore \angle CTB=\angle OTB-\angle OTC=30^\circ$,\\
$\because TC=TB=2$,\\
$\therefore \angle PBT=\angle TCB=\frac{1}{2}\left(180^\circ-\angle CTB\right)=75^\circ$,\\
In $\triangle TCD$ right-angled at $D$, $\angle CTB=30^\circ$, $TC=2$,\\
$\therefore CD=\frac{1}{2}TC=1$, $TD=\sqrt{3}CD=\sqrt{3}$,\\
$\therefore BD=TB-TD=2-\sqrt{3}$,\\
$\therefore CB=\sqrt{CD^{2}+DB^{2}}=\sqrt{1^{2}+(2-\sqrt{3})^{2}}=\sqrt{8-4\sqrt{3}}=\sqrt{8-2\sqrt{12}}=\sqrt{(\sqrt{6}-\sqrt{2})^{2}}=\boxed{\sqrt{6}-\sqrt{2}}$,\\


Question: As shown in the figure, a square frame with a side length of 10 cm (ignoring the thickness of the frame) is pulled into a quadrilateral $ABCD$. If $\angle BAD=60^\circ$, what is the length of $AC$ in cm?

Solution:
\textbf{Solution:} As shown in the figure, construct $DE\perp AB$ intersecting $AB$ at point E, and $CF\perp AB$ at its extension at point F, and connect $AC$,\\
since $AB=BC=CD=AD=10$cm, and $\angle BAD=60^\circ$,\\
thus $DE=CF=AD\times \sin 60^\circ=10\times \frac{\sqrt{3}}{2}=5\sqrt{3}$,\\
$BF=BC\times \cos 60^\circ=10\times \frac{1}{2}=5$,\\
in right triangle $AFC$, $AC=\sqrt{AF^{2}+CF^{2}}$,\\
thus $AC=\sqrt{15^{2}+{(5\sqrt{3})}^{2}}=\boxed{10\sqrt{3}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $\angle B=30^\circ$, and $BC=4\sqrt{3}$. Point P is a movable point on the right-angled side BC (P does not coincide with B or C). Connect AP and rotate line segment AP clockwise by $60^\circ$ around point A to obtain line segment AD. Connect CD. What is the minimum value of line segment CD?

Solution: \textbf{Solution:} Rotate the line segment AC clockwise by $60^\circ$ around point A to get the line segment AE, and connect DE. Then, as point D moves on DE, when CD$\perp$DE, the value of the line segment CD is minimized. Draw CF$\perp$AE at point F through point C. By the property of rotation: $\triangle AED\cong \triangle ACP$, $\therefore \angle AED=\angle ACP=90^\circ$. Since CD$\perp$DE and CF$\perp$AE, $\therefore$ quadrilateral EDCF is a rectangle, $\therefore$ CD=EF. In $\triangle ABC$, $\angle ACB=90^\circ$, $\angle B=30^\circ$, BC=$4\sqrt{3}$, $\therefore$ AC=$BC\tan30^\circ=4\sqrt{3}\times\frac{\sqrt{3}}{3}=4$, $\therefore$ AE=AC=4. In $\triangle ACF$, $\angle AFC=90^\circ$, $\angle FAC=60^\circ$, $\therefore$ AF=$AC\times\cos60^\circ=4\times\frac{1}{2}=2$, $\therefore$ EF=AE−AF=4−2=2, $\therefore$ CD=\boxed{2}.


Question: As shown in the figure, in $\triangle ABC$, $\angle ACB=90^\circ$, $D$ is the midpoint of $AB$, $DE\perp AB$, intersecting $AC$ at $E$. If $\frac{AE}{EC}=\frac{5}{3}$, then what is the value of $\tan\angle A$?

Solution: \textbf{Solution:} As shown in the figure, connect BE. Since $D$ is the midpoint of $AB$ and $DE \perp AB$, it follows that AE = BE. Since $\frac{AE}{EC} = \frac{5}{3}$, we can set AE = BE = 5k and CE = 3k. By the Pythagorean theorem, we get BC = 4k. Therefore, AC = AE + CE = 8k. Therefore, $\tan A = \frac{BC}{AC} = \frac{4k}{8k} = \boxed{\frac{1}{2}}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle ABC=90^\circ$, $BD\perp AC$ at point $D$, $AC=10$, $\cos C=\frac{3}{5}$, then how long is $CD$?

Solution:
Solution: Since $BD\perp AC$ and $\angle ABC=90^\circ$, \\
it follows that $\cos C=\frac{CD}{BC}=\frac{BC}{AC}=\frac{3}{5}$, \\
given that $AC=10$, \\
it follows that $BC=6$, \\
hence, $CD=\frac{3}{5}BC=\boxed{\frac{18}{5}}$


Question: As shown in the figure, in the isosceles right $\triangle ABC$, where $\angle C=90^\circ$, point D lies on AC. If $CD=6$ and $\sin\angle CBD=\frac{3}{5}$, then what is the length of AB?

Solution:
\textbf{Solution:} Given: $\because$ $\angle C=90^\circ$,\\
$\therefore$ $\sin\angle CBD=\frac{CD}{BD}=\frac{3}{5}$,\\
$\because$ $CD=6$,\\
$\therefore$ $\frac{6}{BD}=\frac{3}{5}$, solving this we get: $BD=10$,\\
$\therefore$ $BC=\sqrt{BD^{2}-CD^{2}}=\sqrt{10^{2}-6^{2}}=8$,\\
$\because$ isosceles right $\triangle ABC$,\\
$\therefore$ $AC=BC=8$,\\
$\therefore$ $AB=\sqrt{AC^{2}+BC^{2}}=\sqrt{8^{2}+8^{2}}=\boxed{8\sqrt{2}}$.


Question: As shown in the figure, $PA$ is the tangent line to $\odot O$ at point $A$, and $PO$ intersects $\odot O$ at point $B$. Given $\tan P = \frac{3}{4}$ and $OB = 6$, what is the length of $PB$?

Solution:
\textbf{Solution:} Given that $PA$ is the tangent of circle $O$ with A as the point of tangency,\\
thus $OA\perp AP$,\\
hence $\angle OAP=90^\circ$\\
In $\triangle OAP$,\\
since $\tan P=\frac{OA}{AP}=\frac{3}{4}$ and $OA=OB=6$,\\
we have $\frac{6}{AP}=\frac{3}{4}$,\\
therefore $AP=8$,\\
thus $OP=\sqrt{OA^{2}+AP^{2}}=10$,\\
hence $PB=OP-OB=10-6=\boxed{4}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are at the vertices of a square grid. What is the value of $\sin A$?

Solution:
\textbf{Solution:} Draw line segment DC, \\
from the grid we know that $CD\perp AB$, \\
thus $DC = \sqrt{2}$, $AC = \sqrt{10}$, \\
hence $\sin A = \frac{DC}{AC} = \frac{\sqrt{2}}{\sqrt{10}} = \boxed{\frac{\sqrt{5}}{5}}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are all on the grid points of the square grid paper. What is the value of $\sin C$?

Solution:
\textbf{Solution:} As the figure shows, connect AD.
According to the problem:
\[AC^{2}=1^{2}+7^{2}=50,\]
\[CD^{2}=1^{2}+2^{2}=5,\]
\[AD^{2}=6^{2}+3^{2}=45,\]
\[\therefore AD^{2}+CD^{2}=AC^{2},\]
\[\therefore \triangle ACD \text{ is a right-angled triangle},\]
\[\therefore \angle ADC=90^\circ,\]
In $Rt\triangle ACD$, $AD=3\sqrt{5}$, $AC=5\sqrt{2}$,
\[\therefore \sin C=\frac{AD}{AC}=\frac{3\sqrt{5}}{5\sqrt{2}}=\boxed{\frac{3\sqrt{10}}{10}}.\]


Question: As shown in the figure, the grid is a square grid, and points A, B, C are the intersections of the grid lines. Let $\angle BAC = \alpha$, then what is $\tan\alpha$?

Solution:
\textbf{Solution:} Let the side length of the small square be: 1,\\
then, from the figure we know: $AB=\sqrt{1^2+3^2}=\sqrt{10}$, $BC=\sqrt{1^2+2^2}=\sqrt{5}$, $AC=\sqrt{4^2+3^2}=5$,\\
$S_{\triangle ABC}=3\times 4-1-\frac{1}{2}\times 1\times 2-\frac{1}{2}\times 1\times 3-\frac{1}{2}\times 4\times 3=\frac{5}{2}$,\\
Draw $BD\perp AC$ passing through point $B$, intersecting $AC$ at point $D$,\\
then: $S_{\triangle ABC}=\frac{1}{2}AC\cdot BD=\frac{1}{2}\times 5\cdot BD=\frac{5}{2}$,\\
$\therefore$ $BD=1$,\\
$\therefore$ $AD=\sqrt{AB^2-BD^2}=\sqrt{10-1}=3$,\\
$\therefore$ $\tan\alpha =\frac{BD}{AD}=\boxed{\frac{1}{3}}$.


Question: As shown in the figure, the vertices of $\triangle ABC$ are located on the grid points of a square grid. What is the value of $\tan A$?

Solution:
\textbf{Solution:} As shown in the figure, connect $BD$,\\
due to the characteristics of the grid, we have $BD \perp AC$,\\
$AD = \sqrt{2^2 + 2^2} = 2\sqrt{2}$, $BD = \sqrt{1^2 + 1^2} = \sqrt{2}$,\\
$\therefore \tan A = \frac{BD}{AD} = \frac{\sqrt{2}}{2\sqrt{2}} = \boxed{\frac{1}{2}}$


Question: As shown in the figure, with C as the common vertex, in $\triangle ABC$ and $\triangle CED$, $\angle ACB=\angle CDE=90^\circ$ and $\angle A=\angle DCE=30^\circ$, and with point D on line segment $AB$, what is $\angle ABE$? If $AC=10$ and $CD=9$, what is the length of $BE$?

Solution:
\textbf{Solution:}

Given that $\angle ACB=\angle CDE=90^\circ$ and $\angle A=\angle DCE=30^\circ$,\\
it follows that $\angle DBC=\angle DEC=60^\circ$,\\
hence, points B, C, D, and E are concyclic,\\
therefore $\angle DBE=\angle DCE=30^\circ$,\\
which leads us to conclude that $\angle ABE=\boxed{30^\circ}$,\\
considering that in right triangle $ABC$, $\tan\angle BAC=\frac{BC}{AC}$,\\
thus $BC=AC\cdot\tan30^\circ=10\times \frac{\sqrt{3}}{3}=\frac{10\sqrt{3}}{3}$,\\
similarly, in right triangle $CED$, $\cos\angle DCE=\frac{CD}{CE}$,\\
therefore $CE=\frac{CD}{\cos30^\circ}=\frac{9}{\frac{\sqrt{3}}{2}}=6\sqrt{3}$,\\
given $\angle ABC=60^\circ$ and $\angle ABE=30^\circ$,\\
it follows that $\angle CBE=\angle ABC+\angle ABE=90^\circ$,\\
consequently, $BE=\sqrt{CE^2-BC^2}=\sqrt{108-\frac{300}{9}}=\boxed{\frac{4\sqrt{42}}{3}}$.


Question: In the figure, within $\triangle ABC$, $\angle C=90^\circ$, point $D$ is on $BC$, $AD=BC=5$, and $\cos\angle ADC=\frac{3}{5}$. What is the value of $\tan B$?

Solution:
\textbf{Solution:} Given: $AD=BC=5$, \\
$\cos\angle ADC=\frac{3}{5}$, \\
Hence, $\frac{CD}{AD}=\frac{3}{5}$, \\
Therefore, $CD=3$, \\
By the Pythagorean theorem, we have: $AC=\sqrt{AD^{2}-CD^{2}}=\sqrt{5^{2}-3^{2}}=4$, \\
Hence, $\tan B=\frac{AC}{BC}=\boxed{\frac{4}{5}}$


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $\angle B=30^\circ$. Point $D$ is on side $AB$, and point $E$ is on side $BC$. After folding $\triangle BDE$ along line $DE$, point $B$ falls exactly on point $P$, which is on the extension of line segment $AC$. If $\angle APE=2\angle B$, what is the value of $\frac{BD}{AD}$?

Solution:
\textbf{Solution:} Let's consider $Rt\triangle ABC$ where $\angle C=90^\circ$,\\
$\because \angle B=30^\circ$,\\
$\therefore \angle A=60^\circ$,\\
$\therefore \angle APE=2\angle B=60^\circ$,\\
By folding, it is known that: $PD=BD\text{, }\angle DPE=\angle B=30^\circ$,\\
$\therefore \angle APD=\angle APE−\angle DPE=30^\circ$,\\
$\therefore \angle ADP=60^\circ+30^\circ=90^\circ$,\\
$\therefore \frac{PD}{AD}=\sqrt{3}$,\\
$\therefore \frac{BD}{AD}=\boxed{\sqrt{3}}$.


Question: As shown in the figure, $AB$ is the diameter of the semicircle $O$, and points $E$, $C$, $D$ are three points on the semicircular arc, with $AC \parallel OD$ and $AB \parallel CD$. Given that $AB=12$ and $\angle EAC = 15^\circ$, and connecting $OE$ that intersects $AC$ at point $F$, what is the length of $EF$?

Solution:
\textbf{Solution:} Let's connect OC. \\
Since $AC\parallel OD$, $AB\parallel CD$, and $OA=OD$, \\
it follows that quadrilateral AODC is a rhombus. \\
Thus, $AC=OD$. \\
Given $AB=12$, \\
it follows that $OE=OC=AC=OA=\frac{1}{2}AB=6$. \\
Hence, $\triangle AOC$ is an equilateral triangle. \\
Thus, $\angle AOC=60^\circ$. \\
Since $\angle EAC=15^\circ$, \\
it follows that $\angle EOC=2\angle EAC=30^\circ$. \\
Thus, $OE$ bisects $\angle AOC$. \\
Therefore, $OE\perp AC$. \\
In $\triangle OFC$, $\cos\angle EOC=\frac{OF}{OC}=\frac{\sqrt{3}}{2}$. \\
Thus, $OF=\frac{\sqrt{3}}{2}OC=3\sqrt{3}$. \\
Therefore, $EF=OE-OF=6-3\sqrt{3}$. \\
Hence, the answer is: $EF=\boxed{6-3\sqrt{3}}$.


Question: As shown in the figure, within a square grid with side length 1, points A, B, and C are on grid points, and lines AB and CD intersect at point D. What is the value of $\sin \angle ADC$?

Solution:
\textbf{Solution:} Extend CD to meet another vertex of the square at $E$, and connect BE, as shown in the diagram:\\
From the problem statement, it is known that: $\angle BED=90^\circ$, $\angle ADC=\angle BDE$,\\
Based on the side length of the small square grid and the Pythagorean theorem, it can be derived that: $BE=2\sqrt{2}$, $BD=\sqrt{10}$,\\
$\therefore$ In $\triangle BDE$, $\sin\angle BDE=\frac{BE}{BD}=\frac{2\sqrt{5}}{5}$,\\
$\therefore \sin\angle ADC=\sin\angle BDE=\boxed{\frac{2\sqrt{5}}{5}}$.


Question: As shown in the diagram, $AB=AC=4$ and $\angle BAC=90^\circ$. Let point $M$ be a movable point on line segment $AC$, and connect $BM$. Fold line segment $BA$ along line $BM$ so that point $A$ falls on point $N$. Connect $CN$, and construct an isosceles right-angled triangle $CND$ on the left side (or below) of line $CN$, using $CN$ as the hypotenuse. During the process when point $M$ moves from $A$ to $C$, what is the minimum value of the line segment $CD$? What is the total path length traversed by point $D$ when $M$ moves from point $A$ to $C$?

Solution: \textbf{Solution:} Connect $AC$ and $AD$, with point $N$ moving on a $\frac{1}{4}$ circle centered at $B$ with radius $4$ (from $A$ to $N'$), resulting in $BN=AB=4$. In $\triangle ABC$, we have $AB=AC$ and $\angle BAC=90^\circ$, thus $\angle BCA=45^\circ$ leading to $BC=\sqrt{2}AC=4\sqrt{2}$. When $C$, $N$, and $B$ are collinear, $CN$ is minimized, hence the minimum value of $CN$ is $BC-4=4\sqrt{2}-4$. In the isosceles right triangle $\triangle CDN$, the minimum $CD$ is $\frac{\sqrt{2}}{2}CN=\frac{\sqrt{2}}{2}(4\sqrt{2}-4)=4-2\sqrt{2}$. Since $AB=AC$ and $\angle BCA=90^\circ$, we have $\angle ACB=\angle ABC=45^\circ$. Similarly, $\angle DCN=45^\circ$, hence $\angle ACB=\angle DCN$, leading to $\angle ACB-\angle ACN=\angle DCN-\angle ACN$, which means $\angle BCN=\angle ACD$. Given $\frac{BC}{AC}=\frac{CN}{CD}=\sqrt{2}$, we conclude that $\triangle BCN\sim \triangle ACD$. Therefore, $\frac{AD}{BN}=\frac{AC}{BC}=\frac{\sqrt{2}}{2}$, yielding $AD=\frac{\sqrt{2}}{2}BN=\frac{\sqrt{2}}{2}\times 4=2\sqrt{2}$. Thus, point $D$ moves on a $\frac{1}{4}$ circle centered at $A$ with radius $2\sqrt{2}$. Therefore, as point $M$ moves from $A$ to $C$, point $D$ moves on quarter circle $A$, and since $\frac{1}{4}\times 2\pi \cdot 2\sqrt{2}=\sqrt{2}\pi$, the path length of point $D$ is $\sqrt{2}\pi$. Consequently, the minimum value of $CD$ is $\boxed{4-2\sqrt{2}}$, and the total path length of point $D$ is $\boxed{\sqrt{2}\pi}$.


Question: As shown in the figure, $ \triangle ABC$ is inscribed in a circle $\odot O$ with a radius of 6. Given $\sin\angle A=\frac{2}{3}$, what is the length of $BC$?

Solution:
\textbf{Solution:} Construct the diameter $CD$ of $\odot O$, and connect $BD$, hence $CD=2\times 6=12$.\\
$\therefore$ $\angle CBD=90^\circ$, $\angle D=\angle A$,\\
$\therefore$ $BC=CD\cdot \sin D=CD\cdot \sin A=12\times \frac{2}{3}=\boxed{8}$.


Question: As shown in the figure, rectangle $ABCD$ is folded such that point $D$ falls on the trisection point $G$ of side $AB$, with $BG<AG$. The corresponding point of point $C$ after folding is $H$, and the fold line is $EF$. Connect $BH$. If $\tan\angle AEG = \frac{4}{3}$ and $EF=3\sqrt{5}$, what is the length of $BH$?

Solution:
\textbf{Solution:}

Construct $MN\perp AB$ extending to $N$ on line $AB$, and $FM\perp AD$ intersecting $AD$ at $M$. Draw $DG$ intersecting $EF$ at $l$. Since quadrilateral $ABCD$ is a rectangle, it follows that $\angle ADC=\angle A=\angle ABC=90^\circ$, $AD\parallel BC$, and $CD=AB$. Therefore, $\angle AGE+\angle AEG=90^\circ$. By folding, we obtain $EG=DE$, $\angle HGE=\angle ADC=90^\circ$. $EF$ bisects $DG$ perpendicularly, hence $\angle HGN+\angle AGE=90^\circ$, $\angle FED+\angle ADG=90^\circ$, $HG=CD$. Therefore, $\angle AGE=\angle HGN$. Given that $\tan\angle AEG=\frac{4}{3}$, we can set $AG=4x$ and $AE=3x$, thus $DE=EG=5x$, $BG=2x$, $AD=8x$. $\tan\angle ADG=\frac{1}{2}$. Since $FM\perp AD$, it follows that $\angle FED+\angle EFM=90^\circ$, $FM=AB=AG+BG=6x$. Therefore, $\angle ADG=\angle EFM$, that is, $\tan\angle EFM=\tan\angle ADG=\frac{1}{2}$. Therefore, $ME=3k$, hence $ME^{2}+FM^{2}=EF^{2}$, which implies $(3x)^{2}+(6x)^{2}=(3\sqrt{5})^{2}$. Solving gives $k=1$ or $-1$ (discard), therefore, $BG=2$, $HG=CD=AB=6$. Given that $\tan\angle HGN=\frac{4}{3}$, we can let $HN=4a$ and $NG=3a$. Hence, $HN^{2}+NG^{2}=HG^{2}$, which implies $(4a)^{2}+(3a)^{2}=6^{2}$. Solving gives $a=\frac{6}{5}$, hence $HN=\frac{24}{5}$, $NG=\frac{18}{5}$. Therefore, $NB=NG-BF=\frac{8}{5}$, therefore $BH=\sqrt{HN^{2}+BN^{2}}=\boxed{\frac{8\sqrt{10}}{5}}$.


Question: In the Cartesian coordinate system $xOy$, the graphs of the linear functions $y=kx$ and $y=mx+n$ are shown as in the figure. What is the solution set of the linear inequality in one variable with respect to $x$, $(k-m)x-n>0$?

Solution:
\textbf{Solution:} From $(k-m)x-n>0$, we obtain: $kx>mx+n$,\\
According to the graph, the intersection point of the two functions is $(1, 2)$,\\
Therefore, the solution set of the linear inequality in $x$, $kx>mx+n$, is $x>1$, that is, the solution set of the linear inequality in $x$, $(k-m)x-n>0$, is $\boxed{x>1}$.


Question: The graph of the function $y=|2x+1|-2$ is shown in the figure. What is the range of the function values $y$ when $-2<x<2$?

Solution:
\textbf{Solution:} When $x=2$, we have $y=\left|2x+1\right|-2=3$,\\
since $\left|2x+1\right|\ge 0$,\\
therefore, the minimum value of the function is $-2$,\\
according to the graph, when $-2<x<2$, the range of the function value $y$ is $-2\leq y<3$.\\
Thus, the answer is: $\boxed{-2\leq y<3}$.


Question: As shown in the diagram, it is known that the graphs of the functions $y=2x$ and $y=ax+4$ ($a$ is a constant, and $a\neq 0$) intersect at point $A(1, 2)$. What is the solution set of the inequality $ax+4\ge 2x$ with respect to $x$?

Solution:
\textbf{Solution:} The graphs of the functions $y=2x$ and $y=ax+4$ intersect at point $A\left(1, 2\right)$, which means when $x=1$, $2x=ax+4$. \\
By observing the graph, we can conclude that when $x\le 1$, $2x\le ax+4$\\
Therefore, the answer is $\boxed{x\leq 1}$.


Question: As shown in the figure, the line $y=kx+b$ $(k<0)$ passes through $A(3, 1)$. What is the range of $x$ when $kx+b\le \frac{1}{3}x$?

Solution:
\textbf{Solution:} Since the direct proportion function $y= \frac{1}{3}x$ also passes through point A, \\
it follows that the solution set for $kx+b\leq \frac{1}{3}x$ is $\boxed{x\geq 3}$.


Question: As shown in the figure, in the Cartesian coordinate system, points $A_{1}$, $A_{2}$, $A_{3}$, ..., are all on the positive $x$-axis, and points $B_{1}$, $B_{2}$, $B_{3}$, ..., are all on the line $y=\frac{\sqrt{3}}{3}x$. The triangles $\triangle A_{1}B_{1}A_{2}$, $\triangle A_{2}B_{2}A_{3}$, $\triangle A_{3}B_{3}A_{4}$, ..., are all equilateral triangles, and $OA_{1}=1$. What is the $y$-coordinate of point $B_{6}$?

Solution:
\textbf{Solution:} Let the side length of $\triangle B_{n}A_{n}A_{n+1}$ be $a_{n}$,\\
since points $B_{1}$, $B_{2}$, $B_{3}$, ... are on the line $y = \frac{\sqrt{3}}{3}x$ within the first quadrant,\\
draw $A_{1}N \perp x$ axis from $A_{1}$ to intersect the line $OB_{1}$ at point $N$,\\
since $OA_{1} = 1$,\\
therefore, the $x$-coordinate of point $N$ is $1$, substituting $x=1$ into $y= \frac{\sqrt{3}}{3}x$, we get $y= \frac{\sqrt{3}}{3}$,\\
therefore, the coordinates of point $N$ are $(1, \frac{\sqrt{3}}{3})$,\\
therefore $A_{1}N= \frac{\sqrt{3}}{3}$,\\
in $\triangle NOA_{1}$, $\tan\angle A_{1}ON= \frac{A_{1}N}{A_{1}O} = \frac{\sqrt{3}}{3}$,\\
therefore $\angle A_{1}OB_{1} = 30^\circ$,\\
also, since $\triangle B_{n}A_{n}A_{n+1}$ is an equilateral triangle,\\
therefore $\angle B_{n}A_{n}A_{n+1} = 60^\circ$,\\
therefore $\angle OB_{n}A_{n} = 30^\circ$, $A_{n}B_{n} = OA_{n}$,\\
since $OA_{1}=1$, $a_{1} =1$, $a_{2}=1+1=2= 2a_{1}$, $a_{3}= 1+a_{1} +a_{2}=4= 2a_{2}$, $a_{4}= 1+a_{1}+a_{2}+a_{3}=8= 2a_{3}$, $a_{n+1}= 2a_{n}$, $a_{5}=2a_{4}= 16$, $a_{6}= 2a_{5} = 32$, $a_{7}= 2a_{6}= 64$,\\
$\triangle A_{6}B_{6}A_{7}$ is an equilateral triangle,\\
the coordinates of point $B_{6}$ are $\left(a_{7}- \frac{1}{2} a_{6}, \frac{\sqrt{3}}{3} (a_{7}- \frac{1}{2} a_{6})\right)$,\\
therefore, the coordinates of point $B_{6}$ are $(48,16 \sqrt{3})$,\\
thus, the $y$-coordinate of point $B_{6}$ is $\boxed{16 \sqrt{3}}$.


Question: As shown in the figure, it is known that the graphs of the functions $y=x+b$ and $y=ax+4$ intersect at point $P$. What is the solution set of the inequality $x+b>ax+4$?

Solution:
\textbf{Solution:} Since the intersection point of the graphs of the function $y=x+b$ and $y=ax+4$ has an x-coordinate of $1$, \\
it follows that the solution set for the inequality $x+b>ax+4$ is $\boxed{x>1}$.


Question: As shown in the figure, the line $y=kx+b$ passes through points $A(2,1)$ and $B(-1,-2)$. What is the solution set for the inequality $-2<kx+b<1$?

Solution:
\textbf{Solution:} Because the line $y=kx+b$ passes through points $A(2, 1)$ and $B(-1, -2)$, \\
therefore $\begin{cases}2k+b=1\\ -k+b=-2\end{cases}$, solving this yields $\begin{cases}k=1\\ b=-1\end{cases}$,\\
Because $-2<kx+b<1$,\\
therefore $-2<x-1<1$,\\
therefore $\begin{cases}x-1<1\\ x-1>-2\end{cases}$, solving this gives $-1<x<2$,\\
hence, the answer is $x \in \boxed{(-1, 2)}$.


Question: As shown in the figure, it is known that the graphs of the functions $y=3x+b$ and $y=ax-3$ intersect at point $P(-2, -5)$. Based on the graph, what is the solution set of the inequality $3x+b>ax-3$?

Solution:
\textbf{Solution:} Since the graphs of the functions $y=3x+b$ and $y=ax-3$ intersect at point $P(-2,-5)$,\\
it follows that the solution set of the inequality $3x+b>ax-3$ is $x>-2$,\\
therefore, the answer is: $\boxed{x>-2}$.


Question: The graph of the linear function $y=ax+b$ and the direct proportion function $y=kx$ are shown in the same Cartesian coordinate system as in the figure. What is the solution set of the inequality $ax+b\geq kx$ with respect to $x$?

Solution:
\textbf{Solution:} The intersection point of the two lines is $(-1, 2)$. Given that for $x \geq -1$, the line $y=kx$ lies below the line $y=ax+b$, therefore, the solution set of the inequality $ax+b \geq kx$ is $x \geq -1$.\\
Hence, the solution set is $\boxed{x \geq -1}$.


Question: As shown in the figure, the graph of the linear function $y=-x+1$ intersects with the graph of $y=2x+m$ at point $P\left(n, 2\right)$. What is the solution set of the inequality $-x+1>2x+m>0$ with respect to $x$?

Solution:
Solution: Since the graph of the linear function $y=-x+1$ passes through point $P(-1,2)$,
we have $2=-(-1)+1$, solving this gives $n=-1$,
therefore $P(-1,2)$,
substituting $P(-1,2)$ into $y=2x+m$, we get $2=-2+m$,
solving this gives $m=4$,
thus $y=2x+4$,
when $y=0$, $2x+4=0$, solving this gives $x=-2$,
therefore, the intersection of $y=2x+4$ and the x-axis is $(-2,0)$,
thus, the solution set of the inequality in $x$, $-x+1>2x+m>0$, is $-2<x<-1$.
Therefore, the answer is $\boxed{-2<x<-1}$.


Question: As shown in the figure, the coordinate of point \(A\) is \( \left(2, 3\right) \), the coordinate of point \(B\) is \( \left(4, 0\right) \), and point \(C\) is located in the fourth quadrant. If the quadrilateral formed by the vertices \(O\), \(A\), \(B\), and \(C\) is a rhombus, what are the coordinates of point \(C\)?

Solution:
\textbf{Solution:} Given the property of a diamond that its diagonals are perpendicular and bisect each other, it is known that points $A$ and $C$ are symmetrical about the $x$-axis,\\
since $A(2, 3)$,\\
the symmetry about the $x$-axis keeps the $x$-coordinate the same and changes the sign of the $y$-coordinate,\\
therefore $C(2, -3)$,\\
thus, the answer is: $C=\boxed{(2, -3)}$.


Question: As shown in the diagram, the line $l_{1}: y=x+3$ and the line $l_{2}: y=kx+b$ intersect at a point in the same Cartesian coordinate system. What is the solution set of the inequality $x+3<kx+b$ with respect to $x$?

Solution:
\textbf{Solution:} Since the line $l_{1}: y=x+3$ intersects with the line $l_{2}: y=kx+b$ at the point $(-1,2)$, \\
it follows that when $x<-1$, $x+3<kx+b$, \\
thus, the solution set for the inequality $x+3<kx+b$ with respect to $x$ is $x<-1$. \\
Therefore, the answer is $\boxed{x<-1}$.


Question: As illustrated in the diagram, the line $l_{1}: y=x+2$ intersects the $x$ axis at point $A$ and the $y$ axis at point $B$, while the line $l_{2}: y=4x-4$ intersects the $y$ axis at point $C$ and the $x$ axis at point $D$. The lines $l_{1}$ and $l_{2}$ intersect at point $P$. If there exists a point $Q$ on the $x$ axis such that the quadrilateral formed by vertices $A$, $C$, $P$, $Q$ is a parallelogram, then what are the coordinates of point $Q$?

Solution:
\textbf{Solution:} Given $y=x+2$, when $y=0$, we have $x+2=0$,\\
Solving yields: $x=-2$,\\
$\therefore$ The coordinates of point A are $(-2,0)$,\\
In $y=4x-4$, when $x=0$, we have $y=-4$,\\
$\therefore$ The coordinates of point C are $(0,-4)$,\\
By solving the system of equations $\begin{cases} y=x+2 \\ y=4x-4 \end{cases}$,\\
We find: $\begin{cases} x=2 \\ y=4 \end{cases}$,\\
$\therefore$ The coordinates of point P are $(2,4)$,\\
Let the coordinates of point Q be $(x,0)$,\\
$\because$ Point Q is on the x-axis,\\
$\therefore$ when the quadrilateral with vertices A, C, P, Q is a parallelogram, AQ and PC are diagonals,\\
$\therefore$ $\frac{-2+x}{2}=\frac{2+0}{2}$,\\
Solving yields: $x=4$,\\
$\therefore$ The coordinates of point Q are $\boxed{(4,0)}$.


Question: In the Cartesian coordinate system $xOy$, the graphs of the linear functions $y=kx$ and $y=mx+n$ are shown as in the figure. What is the solution set of the linear inequality $kx>mx+n$ in terms of $x$?

Solution:
\textbf{Solution:} From the graph, it's known that the intersection point of the two functions is $(1,2)$,\\
thus the solution set of the linear inequality in $x$, $kx - n > mx$, is $x > 1$,\\
therefore, the answer is: $\boxed{x > 1}$.


Question: As shown in the figure, in the Cartesian coordinate system, the rectangle $OABC$ has two adjacent sides on the coordinate axes, with vertex $B(6, 4)$. A line passing through point $P(4, m)$ on side $BC$ bisects the area of the rectangle. What is the equation of this line?

Solution:
\textbf{Solution:}

Since the adjacent sides of rectangle OABC are on the coordinate axes, and the vertex B is at (6, 4),\\
therefore, the intersection point of the diagonal lines of the rectangle is at (3, 2),\\
since point P(4, m) is a point on side BC,\\
therefore, P(4, 4),\\
since a line passing through the intersection point of the diagonal lines bisects the rectangle,\\
therefore, let the line passing through P(4, m) and bisecting the rectangle be $y=kx+b$,\\
substituting points (4, 4) and (3, 2) yields $\begin{cases}4k+b=4\\ 3k+b=2\end{cases}$,\\
solving, we get $\begin{cases}k=2\\ b=-4\end{cases}$,\\
therefore, the equation of this line is $y=2x-4$.\\
Hence, the answer is: $\boxed{y=2x-4}$.


Question: As shown in the figure, the point $P(-4, 3)$ is on the graph of the linear function $y = kx + b$ (where $k \neq 0$). Then, what is the solution set of the inequality $kx + b < 3$ with respect to $x$?

Solution:
\textbf{Solution:} When $x > -4$, $y < 3$, i.e., $kx + b < 3$,\\
thus the solution set of the inequality $kx + b < 3$ with respect to $x$ is $x > -4$.\\
Therefore, the answer is: $\boxed{x > -4}$.


Question: As shown in the figure, the line $y=x+2$ intersects with the line $y=kx+6$ at point $P(3, n)$. Then, the system of equations is

Solution:
\textbf{Solution:} Let $\because$ the line $y=x+2$ passes through the point $P(3, n)$,\\
$\therefore n=3+2=5$,\\
$\therefore P(3, 5)$,\\
$\because$ the line $y=x+2$ intersects with the line $y=kx+6$ at point $P$,\\
$\therefore$ the solution to the system of equations $\begin{cases}y=x+2\\ y=kx+6\end{cases}$ is $\boxed{\begin{cases}x=3\\ y=5\end{cases}}$.


Question: As shown in the figure, the graph of the linear function $y=kx+b$ intersects the $x$ axis at point $A(3, 0)$ and intersects the $y$ axis at point $B(0, 4)$. It intersects the graph of the direct proportion function $y=ax$ at point $C$, and the $x$-coordinate of point $C$ is 2. What is the solution set of the inequality $0<kx+b<ax$?

Solution:
\textbf{Solution:}

Substitute $A(3, 0)$ and $B(0, 4)$ into $y=kx+b$ to get:
\[
\begin{cases}
b=4\\
3k+b=0
\end{cases}
\]
Solving this yields:
\[
\begin{cases}
b=4\\
k=-\frac{4}{3}
\end{cases}
\]
Therefore, the equation of the linear function is $y=-\frac{4}{3}x+4$.

Substituting $x=2$ into $y=-\frac{4}{3}x+4$ gives $y=\frac{4}{3}$. Therefore, the coordinates of point $C$ are $\left(2, \frac{4}{3}\right)$.

Substituting $C$ into $y=ax$ gives $a=\frac{2}{3}$. Therefore, the equation of the direct proportion function is $y=\frac{2}{3}x$.

The solution set of the inequality $0<kx+b<ax$ is the same as the solution set of $0<-\frac{4}{3}x+4<\frac{2}{3}x$. From the graph, it is found that the solution set of $0<-\frac{4}{3}x+4<\frac{2}{3}x$ is $2<x<3$. Hence, the answer is $\boxed{2<x<3}$.


Question: The graph of the line $l_{1}\colon y=k_{1}x+b$ and the line $l_{2}\colon y=k_{2}x$ in the same Cartesian coordinate system are shown in the figure. The line $l_{1}\colon y=k_{1}x+b$ intersects the x-axis at the point $(-3, 0)$. What is the solution set of the inequality in terms of x, $k_{2}x<k_{1}x+b<0$?

Solution:
\textbf{Solution:} Given the graph, it is apparent that the intersection point of line $l_{1}$ and line $l_{2}$ is at ($-1$, $-2$), and within line $l_{1}$, $y$ decreases as $x$ increases.\\
Because $y=k_{1}x+b$ intersects the $x$-axis at point ($-3$, $0$), and the solution set of the inequality $k_{2}x<k_{1}x+b$ with respect to $x$ is $x<-1$,\\
Therefore, the solution set for the inequality $k_{2}x<k_{1}x+b<0$ with respect to $x$ is $-3<x<-1$,\\
Thus, the answer is: $\boxed{-3<x<-1}$.


Question: The graph of the linear function $y = ax + b$ (where $a$ and $b$ are constants, with $a \neq 0$) is shown as in the diagram. What is the solution to the equation $ax + b = 0$?

Solution:
\textbf{Solution:} Method 1: It can be seen from the graph that the coordinates of the intersection points of the line with the coordinate axes are $(0, -1)$ and $(2, 0)$,\\
substituting $(0, -1)$ and $(2, 0)$ into the linear function $y=ax+b$ yields: $\left\{\begin{array}{l}0=2a+b\\ -1=b\end{array}\right.$,\\
solving this gives $\left\{\begin{array}{l}a=\frac{1}{2}\\ b=-1\end{array}\right.$,\\
$\therefore \frac{1}{2}x-1=0$, solving for $x$ gives $x=2$,\\
$\therefore$ the solution of the equation $ax+b=0$ is $x=\boxed{2}$;\\
Method 2: According to the fact that the abscissa of the intersection point of the linear function $y=ax+b$ with the x-axis is the solution of the linear equation $ax+b=0$, it is known that the solution is $x=\boxed{2}$.


Question: As shown in the figure, the line $l_1: y=kx+b \, (k \neq 0)$ and $l_2: y=mx+n \, (m \neq 0)$ intersect the x-axis at points $A(-4, 0)$ and $B(2, 0)$, respectively. What is the solution set of the system of inequalities $\begin{cases} kx+b>0 \\ mx+n>0 \end{cases}$?

Solution:
Solution: From the graph, we know

$\because$ $kx+b>0$

$\therefore$ $x>-4$

$\because$ $mx+n>0$

$\therefore$ $x<2$

$\therefore$ the solution set of the system of inequalities $\begin{cases}kx+b>0\\ mx+n>0\end{cases}$ is $-4<x<2$

Therefore, the answer is: $\boxed{-4<x<2}$


Question: In the same Cartesian coordinate system, the graphs of the functions $y_1=kx+b$ and $y_2=mx+n$ are shown in the figure. What is the solution set of the inequality $kx+b\geq mx+n$ with respect to $x$?

Solution: \textbf{Solution:} The solution set of the inequality $kx+b \geq mx+n$ is: $\boxed{x \geq 2}$.


Question: As shown in the figure, the graph of the function $y=kx+b$ ($k\neq 0$) passes through the point $(1,2)$. What is the solution set of the inequality $kx+b>2$?

Solution:
\textbf{Solution:} From the graph, it can be inferred that when $x>1$, $kx+b>2$,\\
thus, the solution set of the inequality $kx+b>2$ is $x>1$,\\
therefore, the answer is $\boxed{x>1}$.


Question: The graph of the linear function $y=kx+b$ (where $k$ and $b$ are constants) is shown in the figure. What is the solution set of the linear inequality in one variable $kx+b\ge 0$ with respect to $x$?

Solution:
\textbf{Solution:} When $x\leq 3$, $y\geq 0$, \\
hence the solution set of the linear inequality in one variable $kx+b\geq 0$ with respect to $x$ is $\boxed{x\leq 3}$.


Question: It is known that the graph of the linear function $y=kx+b$ is as shown in the figure. What is the range of $y$ when $x<1$?

Solution:
\textbf{Solution:} \\
Consider the linear function $y=kx+b$. Its graph intersects the $y$-axis at $(0, -4)$, \\
$\therefore b=-4$, and intersects the $x$-axis at $(2, 0)$, \\
$\therefore 0=2k-4$, \\
$\therefore k=2$, \\
$\therefore y=kx+b=2x-4$, \\
$\therefore x=(y+4)/2<1$, \\
$\therefore \boxed{y<-2}$.


Question: As shown in the figure, in the Cartesian coordinate system $xOy$, the line $y=2x-1$ intersects the x-axis and y-axis at point A and B, respectively. Given that $\angle ABC=45^\circ$, what is the equation of line BC?

Solution:
\textbf{Solution:} Since the graph of the linear function $y=2x-1$ intersects the $x$ and $y$ axes at points $A$ and $B$, respectively, \\
let $x=0$, we get $y=-1$, and let $y=0$, we get $x=\frac{1}{2}$, \\
thus, point $A\left(\frac{1}{2},0\right)$, point $B\left(0,-1\right)$, \\
hence, OA$=\frac{1}{2}$, OB$=1$, \\
As shown in the figure, draw $AF\perp AB$ meeting $BC$ at $F$, and draw $FE\perp x$ axis at $E$, \\
since $\angle ABC=45^\circ$, \\
thus, $\triangle ABF$ is an isosceles right triangle, \\
therefore, AB$=$AF, \\
since $\angle OAB+\angle ABO=\angle OAB+\angle EAF=90^\circ$, \\
thus, $\angle ABO=\angle EAF$, \\
therefore, $\triangle ABO\cong\triangle FAE$ (AAS), \\
therefore, AE$=$OB$=1$, EF$=$OA$=\frac{1}{2}$, \\
thus, point $F\left(\frac{3}{2},-\frac{1}{2}\right)$, \\
Let the equation of the line $BC$ be: $y=kx+b$, \\
$\left\{\begin{array}{l} \frac{3}{2}k+b=-\frac{1}{2}\\ b=-1\end{array}\right.$, solving, we get: $\left\{\begin{array}{l}k=\frac{1}{3}\\ b=-1\end{array}\right.$, \\
therefore, the equation of the line $BC$ is: $\boxed{y=\frac{1}{3}x-1}$.


Question: As shown in the figure, given the linear function $y=kx+b(k\ne 0)$ whose graph is illustrated, what is the solution set of the inequality $kx+b>0$?

Solution:
\textbf{Solution:} When $x<1$, we have $y>0$,\\
thus the solution set for the inequality $kx+b>0$ is $x<1$.\\
Therefore, the answer is $\boxed{x<1}$.


Question: As shown in the figure, $AB$ is a chord of the circle $\odot O$, $C$ is the midpoint of $AB$, and $OC$ intersects the arc $\overset{\frown}{AB}$ at point $D$. If $AB=8$ cm and $CD=2$ cm, what is the radius of the circle $\odot O$ in centimeters?

Solution: \textbf{Solution:} Draw line OB. Since point C is the midpoint of arc $\overset{\frown}{AB}$, it follows that OC$\perp$AB and BD=$\frac{1}{2}$AB=4. Let the radius of the circle be r, then OD=r−2. In $\triangle$OBD, $OD^{2}$+$BD^{2}$=$OB^{2}$. Therefore, $(r−2)^{2}$+$4^{2}$=$r^{2}$. Solving this equation, we find $r=\boxed{5}$.


Question: As shown in the diagram, within circle $O$, $OC \perp AB$ at point $C$. If the radius of circle $O$ is $10$ and $AB=16$, then what is the length of $OC$?

Solution:
\textbf{Solution:} Connect OB, \\
$\because$OC$\perp$AB, \\
$\therefore \angle$OCB$=90^\circ$, BC$= \frac{1}{2}$AB$= \frac{1}{2} \times 16=8$, \\
$\therefore OC=\sqrt{OB^{2}-BC^{2}}=\sqrt{10^{2}-8^{2}}=\boxed{6}$.


Question: As shown in the figure, in circle $O$, $\angle ACB=30^\circ$, what is the measure of $\angle AOB$ in degrees?

Solution:
\textbf{Solution:} Given that in circle $\odot O$, $\angle ACB=30^\circ$,\\
therefore, $\angle AOB=2\angle ACB=60^\circ$,\\
hence, the answer is: $\boxed{60^\circ}$.


Question: As shown in the figure, given that the circumference of $\odot O$ is $4\pi$ and the length of $\overset{\frown}{AB}$ is $\pi$, what is the area of the shaded region in the figure?

Solution:
\textbf{Solution:} Since the circumference of circle O is $4\pi$,\\
therefore, the diameter of circle O is 4,\\
thus, the radius of circle O is 2,\\
since the length of arc AB is $\pi$,\\
therefore, the length of arc AB is $\frac{1}{4}$ of the circumference of circle O,\\
hence, $\angle AOB=90^\circ$,\\
therefore, the area of the shaded region $= \frac{90^\circ \cdot \pi \times 2^2}{360} - \frac{1}{2} \times 2 \times 2 = \boxed{\pi-2}$.


Question: As shown in the figure, $AB$ is the diameter of circle $O$, points $C$ and $D$ are on circle $O$, $\angle D=60^\circ$, and $AB=10$. What is the length of $AC$?

Solution:
\textbf{Solution:} Connect $BC$, as shown in the figure,\\
$\because$ $AB$ is the diameter of $\odot O$, and $AB=10$,\\
$\therefore$ $\angle ACB=90^\circ$,\\
$\because$ $\angle D=60^\circ$,\\
$\therefore$ $\angle ABC=60^\circ$,\\
$\therefore$ $\angle CAB=180^\circ-\angle ABC-\angle ACB=30^\circ$,\\
$\therefore$ in $\triangle ABC$, we have: $BC=\frac{1}{2}AB=5$,\\
$\therefore$ $AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{10^{2}-5^{2}}=\boxed{5\sqrt{3}}$


Question: As shown in the figure, a turning point of a road is a section of a circular arc (the arc $ \overset{\frown}{AB}$ in the figure), where point O is the center of this arc, $ \angle AOB=80^\circ$. Point C is on segment AB, and OC is perpendicular to AB at D. Given that $AB=300m$ and $CD=50m$, what is the length of the curved road AB in meters?

Solution:
\textbf{Solution:} Let the radius be $r$,\\
then $OD=r-CD=(r-50)m$,\\
$\because$ $OC\perp AB$,\\
$\therefore$ $AD=BD=\frac{1}{2}AB=\frac{1}{2}\times 300=150m$,\\
In $\triangle AOD$, $AO^2=AD^2+OD^2$,\\
i.e., $r^2=150^2+(r-50)^2$,\\
solving this yields $r=250m$,\\
the radius of this curved segment is $250m$.\\
$\because$ $\angle AOB=80^\circ$,\\
$\therefore$ $\overset{\frown}{AB}=\frac{80\pi \times 250}{180}=\frac{1000\pi}{9}m$.\\
Thus, the length of this curved segment AB is $\boxed{\frac{1000\pi}{9}m}$.


Question: As shown in the figure, in $\triangle ABC$, $\angle C=90^\circ$, $AC=12$, and $BC=5$, what is the distance between the centroid $P$ and the circumcentre $Q$ of $\triangle ABC$?

Solution:
Solution: According to the given information, points C, P, and Q are collinear.

In $\triangle ABC$, $\angle C=90^\circ$, $AC=12$, $BC=5$,

$\therefore AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=13$,

$\because$ the circumcenter of $\triangle ABC$ is $Q$,

$\therefore Q$ is the midpoint of the hypotenuse $AB$,

$\therefore CQ=\frac{1}{2}AB=\frac{13}{2}$,

$\because$ the centroid of $\triangle ABC$ is $P$,

$\therefore PQ=\frac{1}{3}CQ=\boxed{\frac{13}{6}}$.


Question: As shown in the figure, $AB$ is a chord of the circle $\odot O$, C is the midpoint of the arc $AB$, and $OC$ intersects $AB$ at point D. If $AB=8$ cm and $CD=2$ cm, what is the radius of $\odot O$ in cm?

Solution:
\textbf{Solution:} As shown in the diagram below, connect OA,

$\because$ C is the midpoint of arc $AB$,

$\therefore$ $CD\perp AB$, and $CD$ bisects $AB$,

Let the radius of $\odot O$ be $r$,

$\because$ $AB=8$cm, $CD=2$cm,

$\therefore$ $OD=r-2$, $AD=4$,

$\because$ $AO^{2}=AD^{2}+OD^{2}$,

$\therefore$ $r^{2}=4^{2}+(r-2)^{2}$,

$\therefore$ $r^{2}=16+r^{2}-4r+4$,

$\therefore$ $r=\boxed{5}$cm.


Question: As shown in the figure, $AB$ is the tangent to the circle $\odot O$ with point A as the point of tangency. The line $OB$ intersects the circle $\odot O$ at point C. Point D is on the circle $\odot O$. The lines $AD$, $CD$, and $OA$ are connected. If $\angle ABO = 20^\circ$, what is the degree measure of $\angle ADC$?

Solution:
\textbf{Solution:} Given that $AB$ is the tangent to the circle $O$,\\
Therefore, $OA \perp AB$,\\
Therefore, $\angle OAB = 90^\circ$,\\
Given that $\angle B = 20^\circ$,\\
Therefore, $\angle O = 90^\circ - 20^\circ = 70^\circ$,\\
Therefore, $\angle ADC = \frac{1}{2}\angle O = \frac{1}{2} \times 70^\circ = \boxed{35^\circ}$.


Question: As shown in the figure, in the Cartesian coordinate system, with $M(2, 3)$ as the center of the circle and $AB$ as its diameter, the circle is tangent to the x-axis and intersects the y-axis at points A and C. What is the length of $AC$?

Solution:
\textbf{Solution:} As shown in the figure, connect $BC$, and assume the circle is tangent to the x-axis at point $D$, connect $MD$ and it intersects $BC$ at point $E$, so $MD\perp x$ axis,\\
$\because AB$ is the diameter,\\
$\therefore \angle ACB=\angle AOD=90^\circ$,\\
$\therefore BC\parallel x$ axis,\\
$\therefore BC\perp MD$,\\
$\because M\left(2,3\right)$,\\
$\therefore MB=MD=3$, $CE=EB=2$,\\
$\therefore CB=4$, $AB=2BM=6$,\\
$\therefore AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{6^{2}-4^{2}}=\boxed{2\sqrt{5}}$,


Question: As shown in the figure, points $A, B, C$ lie on circle $\odot O$, and $\angle ACB = 40^\circ$. What is the degree measure of arc $AB$?

Solution:
\textbf{Solution:} Since $\angle AOB$ and $\angle ACB$ both subtend $\overset{\frown}{AB}$,\\
it follows that $\angle AOB=2\angle ACB=2\times 40^\circ=80^\circ$,\\
thus, the measure of arc $AB$ is $\boxed{80^\circ}$.


Question: As shown in the figure, it is known that the radius of $\odot O$ is 5, $AB\perp CD$, with the foot of the perpendicular as $P$, and $AB=CD=8$, then what is the length of $OP$?

Solution:
\textbf{Solution:} Connect $OB$, draw $OE\perp AB$ at $E$, and $OF\perp CD$ at $F$,\\
then $BE=\frac{1}{2}AB=4$, the quadrilateral $PEOF$ is a rectangle,\\
since $AB=CD$, $OE\perp AB$, $OF\perp CD$,\\
thus $OE=OF$,\\
therefore, the rectangle $PEOF$ is a square,\\
thus $OE=PE$,\\
in $\triangle OEB$, $OE=\sqrt{OB^{2}-BE^{2}}=3$,\\
thus $OP=\sqrt{OE^{2}+PE^{2}}=\boxed{3\sqrt{2}}$.


Question: As shown in the figure, points C and D lie on a circle with diameter AB, where AB = 2, and $\angle ACD = 30^\circ$. What is the length of AD?

Solution:
\textbf{Solution:} Since AB is the diameter,\\
it follows that $\angle ADB = 90^\circ$,\\
and since $\angle B = \angle ACD = 30^\circ$,\\
it implies that AD = $\frac{1}{2}$AB = $\frac{1}{2} \times 2 =$ \boxed{1}.


Question: As shown in the figure, a circle with center at the origin $O$ passes through point $A(4, 0)$. Inside the circle, there is a fixed point $B(-1, 2)$. A line through point $B$ intersects the circle at points $M$ and $N$. What is the minimum value of $MN$?

Solution:
\textbf{Solution:} As shown in the diagram, connect $OB$ and $OM$. It is known that when $MN\perp OB$, $MN$ is minimized,\\
$\because B(-1,2)$,\\
$\therefore OB^2=1^2+2^2=5$,\\
$\because OM=OA=4$,\\
$\therefore BM=\sqrt{OM^2-OB^2}=\sqrt{16-5}=\sqrt{11}$,\\
$\because MN\perp OB$,\\
$\therefore MN=2BM=2\sqrt{11}$,\\
$\therefore$ The minimum value of $MN$ is $\boxed{2\sqrt{11}}$.