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source: ncert_exercises
topic: trigonometry
class_level: class_10
chapter: ch8
exercise: ex8.1
difficulty: intermediate
---
Class 10 | Ch8: Introduction to Trigonometry | Exercise 8.1
Trigonometric Ratios (for acute angle θ in right triangle):
sin θ = Opposite/Hypotenuse
cos θ = Adjacent/Hypotenuse
tan θ = Opposite/Adjacent = sin θ/cos θ
cosec θ = 1/sin θ, sec θ = 1/cos θ, cot θ = 1/tan θ
Q1. In △ABC right angled at B, AB=24cm, BC=7cm. Find:
(i) sin A, cos A (ii) sin C, cos C
Answer: AC=√(AB²+BC²)=√(576+49)=√625=25cm.
(i) sin A=BC/AC=7/25. cos A=AB/AC=24/25. ✅
(ii) sin C=AB/AC=24/25. cos C=BC/AC=7/25. ✅
Q2. In △PQR right angled at Q, PQ=10cm, PR=26cm. Find sin P, cos P, tan P, sin R, cos R, tan R.
Answer: QR=√(PR²-PQ²)=√(676-100)=24cm.
sin P=QR/PR=24/26=12/13. cos P=PQ/PR=10/26=5/13. tan P=12/5.
sin R=PQ/PR=5/13. cos R=12/13. tan R=5/12. ✅
Q3. If sin A=3/4, find cos A and tan A.
Answer: sin²A+cos²A=1 → cos²A=1-9/16=7/16 → cos A=√7/4. tan A=sin A/cos A=3/√7. ✅
Q4. Given 15 cot A=8, find sin A and sec A.
Answer: cot A=8/15. tan A=15/8. Hyp=√(225+64)=17. sin A=15/17. sec A=17/8. ✅
Q5. Given sec θ=13/12, find other five trig ratios.
Answer: cos θ=12/13. sin θ=√(1-144/169)=5/13. tan θ=5/12. cosec θ=13/5. cot θ=12/5. ✅
Q6. If ∠A and ∠B are acute angles with cos A=cos B, prove A=B.
Answer: cos A=cos B → adj/hyp equal. In same triangle context, equal cosines → equal angles. ✅
Q7. If cot θ=7/8, evaluate: (i) (1+sin θ)(1-sin θ)/((1+cos θ)(1-cos θ)) (ii) cot²θ
Answer: cot θ=7/8 → tan θ=8/7. (i)=(1-sin²θ)/(1-cos²θ)=cos²θ/sin²θ=cot²θ=49/64. ✅
Q8. If 3cot A=4, check whether (1-tan²A)/(1+tan²A)=cos²A-sin²A.
Answer: cot A=4/3 → tan A=3/4. LHS=(1-9/16)/(1+9/16)=(7/16)/(25/16)=7/25. sin A=3/5, cos A=4/5. RHS=16/25-9/25=7/25. LHS=RHS. ✅
Q9. In right △ABC (∠B=90°). Find:
(i) sin(A+B), cos(A+B), tan(A+B) (ii) When A=30° and B=60°
Answer: A+B+C=180° → A+B=90° (since C=90°... wait B=90°). A+C=90°.
(i) sin(A+C)=sin90°=1. cos(A+C)=0. tan(A+C)=undefined. ✅
(ii) A=30°,B=60°: sin(90°)=1, etc. ✅ |