Chapter 8: Introduction to Trigonometry Exercise Questions & Answers – NCERT Class 10 Maths
1In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
Answer & Reference:
Using Pythagoras Theorem: AC² = AB² + BC² = 24² + 7² = 576 + 49 = 625 ⇒ AC = 25 cm.
(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.
(Page 121, Exercise 8.1 Q1)
2In Fig. 8.13, find tan P – cot R.
Answer & Reference:
Assuming standard right triangle with PQ = 12, QR = 5, PR = 13 (as per common NCERT figure):
tan P = QR/PQ = 5/12, cot R = QR/PQ = 5/12 ⇒ tan P – cot R = 0.
(Page 121, Exercise 8.1 Q2)
3If sin A = 3/4, calculate cos A and tan A.
Answer & Reference:
Let BC = 3k, AC = 4k. Then AB = √(AC² – BC²) = √(16k² – 9k²) = √7k.
cos A = AB/AC = √7/4, tan A = BC/AB = 3/√7.
(Page 121, Exercise 8.1 Q3)
4Given 15 cot A = 8, find sin A and sec A.
Answer & Reference:
cot A = 8/15 ⇒ AB = 8k, BC = 15k ⇒ AC = √(64k² + 225k²) = 17k.
sin A = BC/AC = 15/17, sec A = AC/AB = 17/8.
(Page 121, Exercise 8.1 Q4)
5Given sec θ = 13/12, calculate all other trigonometric ratios.
Answer & Reference:
sec θ = 13/12 ⇒ AC = 13k, AB = 12k ⇒ BC = √(169k² – 144k²) = 5k.
sin θ = 5/13, cos θ = 12/13, tan θ = 5/12, cosec θ = 13/5, cot θ = 12/5.
(Page 121, Exercise 8.1 Q5)
6If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer & Reference:
In two right triangles, if cos A = cos B ⇒ adjacent/hypotenuse equal ⇒ triangles similar ⇒ angles equal.
(Page 121, Exercise 8.1 Q6; similar to Example 2, Page 118)
7If cot θ = 7/8, evaluate: (i) (1+sinθ)(1–sinθ)/(1+cosθ)(1–cosθ) (ii) cot²θ
Answer & Reference:
cot θ = 7/8 ⇒ AB = 7k, BC = 8k ⇒ AC = √(49k² + 64k²) = √113k.
sin θ = 8/√113, cos θ = 7/√113.
(i) (1 – sin²θ)/(1 – cos²θ) = cos²θ/sin²θ = cot²θ = (7/8)² = 49/64.
(ii) cot²θ = 49/64.
(Page 121, Exercise 8.1 Q7)
8If 3 cot A = 4, check whether (1 – tan²A)/(1 + tan²A) = cos²A – sin²A or not.
Answer & Reference:
cot A = 4/3 ⇒ tan A = 3/4. LHS = (1 – 9/16)/(1 + 9/16) = (7/16)/(25/16) = 7/25.
cos A = 4/5, sin A = 3/5 ⇒ RHS = 16/25 – 9/25 = 7/25. Hence, LHS = RHS.
(Page 121, Exercise 8.1 Q8)
9In triangle ABC, right-angled at B, if tan A = 1/√3, find: (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C
10In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.
Answer & Reference:
Let QR = x ⇒ PR = 25 – x. By Pythagoras: (25 – x)² = x² + 25 ⇒ 625 – 50x + x² = x² + 25 ⇒ x = 12.
So QR = 12, PR = 13. sin P = 12/13, cos P = 5/13, tan P = 12/5.
(Page 121, Exercise 8.1 Q10; similar to Example 5, Page 120)
11State whether true or false: (i) tan A < 1 always (ii) sec A = 12/5 possible (iii) cos A = cosec A (iv) cot A = cot × A (v) sin θ = 4/3 possible
Answer & Reference:
(i) False – tan 60° = √3 > 1.
(ii) True – sec A ≥ 1 for 0° ≤ A < 90°.
(iii) False – cos A ≠ cosec A.
(iv) False – cot A is a ratio, not product.
(v) False – sin θ ≤ 1 always.
(Page 121, Exercise 8.1 Q11)
TTrigonometric Ratios of Specific Angles (Table 8.1)