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source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch6
exercise: ex6.1
difficulty: intermediate
---
Class 9 | Ch6: Lines and Angles | Exercise 6.1
Key Theorems:
T1: If a ray stands on a line → adjacent angles sum = 180° (Linear pair axiom).
T2: If two lines intersect → vertically opposite angles are equal.
Angle types: Acute (<90°), Right (=90°), Obtuse (90°-180°), Straight (=180°), Reflex (>180°).
Complementary: sum=90°. Supplementary: sum=180°. Linear pair: adjacent + supplementary.
Q1. In the figure, lines AB and CD intersect at O. If ∠AOC+∠BOE=70° and ∠BOD=40°, find ∠BOE and reflex ∠COE.
Answer: ∠AOC=∠BOD=40° (vertically opposite). ∠AOC+∠BOE=70° → 40°+∠BOE=70° → ∠BOE=30°. ∠COE=180°-∠AOC-∠BOE=180°-40°-30°=110° (AOB is straight line). Reflex ∠COE=360°-110°=250°. ✅
Q2. In the figure, lines XY and MN intersect at O. If ∠POY=90° and a:b=2:3, find c.
Answer: ∠POY=90°, so ∠XOM+∠MOY=90° (since POY=90° and XOP is straight). a+b=90°. a:b=2:3 → a=36°, b=54°. c=∠YON=b=54° (vertically opposite)? No: c+b=180° (linear pair on line MN). c=180°-54°=126°. ✅
Q3. In the figure, ∠PQR=∠PRQ. Prove ∠PQS=∠PRT.
Answer: ∠PQR+∠PQS=180° (linear pair at Q on line ST). ∠PRQ+∠PRT=180° (linear pair at R on line ST). Since ∠PQR=∠PRQ → ∠PQS=∠PRT. ✅ Proved.
Q4. In the figure, if x+y=w+z, prove that AOB is a straight line.
Answer: All angles at O: x+y+w+z=360° (angles around a point). x+y=w+z (given). So 2(x+y)=360° → x+y=180°. Since AOC and BOC are adjacent and sum=180°, AOB is a straight line. ✅
Q5. In the figure, POQ is a line. Ray OR is perpendicular to PQ. OS is another ray. Prove ∠ROS = (1/2)|∠QOS - ∠POS|.
Answer: OR⊥PQ → ∠ROP=∠ROQ=90°. ∠QOS+∠ROS=∠ROQ=90° → ∠ROS=90°-∠QOS. ∠POS=∠POR+∠ROS=90°+∠ROS. |∠QOS-∠POS|=|(90°-∠ROS)-(90°+∠ROS)|=|-2∠ROS|=2∠ROS. So ∠ROS=(1/2)|∠QOS-∠POS|. ✅ Proved.
Q6. It is given that ∠XYZ=64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ZYP, find ∠XYQ and reflex ∠QYP.
Answer: ∠ZYP=180°-∠XYZ=180°-64°=116° (linear pair). YQ bisects ∠ZYP → ∠QYP=∠QYZ=58°. ∠XYQ=∠XYZ+∠ZYQ=64°+58°=122°. Reflex ∠QYP=360°-58°=302°. ✅ |