Class 10 Maths MCQs for Chapter 8 Introduction to Trigonometry with Answers PDF

MCQs on Introduction to Trigonometry Class 10 are available in this Maths article along with a free PDF for offline practice. These multiple‑choice questions help students practise the key concepts from Chapter 8 of the CBSE Maths syllabus in an exam‑oriented format. The MCQs with answers and detailed solutions, prepared by our subject experts, cover trigonometric ratios, ratios of complementary angles, trigonometric identities, and their applications to strengthen conceptual understanding and improve problem‑solving skills. By practising MCQs on Introduction to Trigonometry, students can improve accuracy, understand formulas better, and build confidence for board exams.

MCQS on Chapter 8: Introduction to Trigonometry for Class 10 With Answers

Question 1: In △ABC, right-angled at B, AB = 24 cm, BC = 7 cm. The values of sin A and cos A are:

(a) 7/25 and 25/24

(b) 7/25 and 24/25

(c) 24/25 and 7/25

(d) 25/7 and 25/24

Answer: (b) 7/25 and 24/25

Explanation: AC = √(24² + 7²) = √(576 + 49) = √625 = 25 cm.

sin A = BC/AC = 7/25; cos A = AB/AC = 24/25.

Question 2: Given tan A = 4/3. The value of sin A is:

(a) 3/5

(b) 4/5

(c) 4/3

(d) 3/4

Answer: (b) 4/5

Explanation: tan A = 4/3, so BC = 4k, AB = 3k. By Pythagoras: AC = √(16k² + 9k²) = 5k.

sin A = BC/AC = 4k/5k = 4/5.

Question 3: If cos X = a/b, then sin X equals:

(a) (b − a)/b

(b) (b² − a²)/b

(c) √(b² − a²)/b

(d) √(b − a)/b

Answer: (c) √(b² − a²)/b

Explanation: sin²X + cos²X = 1 ⇒ sin²X = 1 − a²/b² = (b² − a²)/b². So sin X = √(b² − a²)/b.

Question 4: In △OPQ, right-angled at P, OP = 7 cm and OQ − PQ = 1 cm. The value of sin Q is:

(a) 24/25

(b) 7/25

(c) 7/24

(d) 1/25

Answer: (b) 7/25

Explanation: Let PQ = x, then OQ = x + 1. OQ² = OP² + PQ² ⇒ (x+1)² = 49 + x² ⇒ 2x + 1 = 49 ⇒ x = 24. OQ = 25.

sin Q = OP/OQ = 7/25. 

Question 5: The value of sin 60° cos 30° + sin 30° cos 60° is:

(a) 0

(b) 1

(c) 1/2

(d) √3/2

Answer: (b) 1

Explanation: sin 60° cos 30° + sin 30° cos 60° = (√3/2)(√3/2) + (1/2)(1/2) = 3/4 + 1/4 = 1.

Question 6: (sin 30° + cos 60°) − (sin 60° + cos 30°) is equal to:

(a) 0

(b) 1 + 2√3

(c) 1 − √3

(d) 1 + √3

Answer: (c) 1 − √3

Explanation: (sin 30° + cos 60°) − (sin 60° + cos 30°) = (1/2 + 1/2) − (√3/2 + √3/2) = 1 − 2(√3/2) = 1 − √3.

Question 7: sin 2A = 2 sin A is true when A =

(a) 30°

(b) 45°

(c) 0°

(d) 60°

Answer: (c) 0°

Explanation: At A = 0°: sin 2A = sin 0° = 0; 2 sin A = 2(0) = 0. 

It holds only at A = 0°.

Question 8: 2 tan 30° / (1 + tan²30°) equals:

(a) sin 60°

(b) cos 60°

(c) tan 60°

(d) sin 30°

Answer: (a) sin 60°

Explanation: tan 30° = 1/√3. Numerator = 2/√3. Denominator = 1 + 1/3 = 4/3.

Result = (2/√3) ÷ (4/3) = (2/√3) × (3/4) = 6/(4√3) = 3/(2√3) = √3/2 = sin 60°.

Question 9: If sin (A − B) = 1/2 and cos (A + B) = 1/2, 0° < A + B ≤ 90°, A > B, then A and B are:

(a) A = 30°, B = 45°

(b) A = 60°, B = 30°

(c) A = 45°, B = 15°

(d) A = 30°, B = 15°

Answer: (c) A = 45°, B = 15°

Explanation: sin(A − B) = 1/2 ⇒ A − B = 30°. cos(A + B) = 1/2 ⇒ A + B = 60°.

Adding: 2A = 90° ⇒ A = 45°, B = 15°.

Question 10: The value of (tan 1° · tan 2° · tan 3° · … · tan 89°) is:

(a) 0

(b) 1

(c) 2

(d) Not defined

Answer: (b) 1

Explanation: Pair each tan k° with tan(90° − k°) = cot k°. Since tan k° × cot k° = 1, all pairs multiply to 1. tan 45° = 1. Overall product = 1.

Question 11: 9 sec²A − 9 tan²A equals:

(a) 1

(b) 9

(c) 8

(d) 0

Answer: (b) 9

Explanation: 9 sec²A − 9 tan²A = 9(sec²A − tan²A) = 9 × 1 = 9. (Identity: sec²A − tan²A = 1)

Question 11: (1 + tan θ + sec θ)(1 + cot θ − cosec θ) equals:

(a) 0

(b) 1

(c) 2

(d) −1

Answer: (c) 2

Explanation: Substitute sin and cos. (1 + sin/cos + 1/cos)(1 + cos/sin − 1/sin)

= [(cos + sin + 1)/cos] × [(sin + cos − 1)/sin]

= [(cos + sin)² − 1] / (sin·cos) = [1 + 2sin·cos − 1] / (sin·cos) = 2sin·cos / (sin·cos) = 2. 

Question 12: The value of cosec [75° + θ] − sec [15° − θ] − tan [55° + θ] + cot [35° − θ] is:

(a) −1

(b) 0

(c) 1

(d) 3/2

Answer: (b) 0

Explanation: cosec(75° + θ) = cosec[90° − (15° − θ)] = sec(15° − θ). So first two terms cancel.

cot(35° − θ) = cot[90° − (55° + θ)] = tan(55° + θ). So last two terms cancel. Total = 0.

Question 13: If tan(A + B) = √3 and tan(A − B) = 1/√3, with 0° < A + B ≤ 90° and A > B, find A.

(a) 30°

(b) 60°

(c) 45°

(d) 15°

Answer: (c) 45°

Explanation: tan(A + B) = √3 ⇒ A + B = 60°. tan(A − B) = 1/√3 ⇒ A − B = 30°.

Adding: 2A = 90° ⇒ A = 45°, B = 15°.

Question 14: If cos(α + β) = 0, then sin(α − β) can be reduced to:

(a) cos β

(b) cos 2β

(c) sin α

(d) sin 2α

Answer: (b) cos 2β

Explanation: cos(α + β) = 0 ⇒ α + β = 90° ⇒ α = 90° − β.

sin(α − β) = sin(90° − β − β) = sin(90° − 2β) = cos 2β.

Question 15: Which of the following is false?

(a) The value of tan A is always less than 1

(b) The value of sin A never exceeds 1

(c) cot A is not defined for A = 0°

(d) cos A decreases as A increases from 0° to 90°

Answer: (a)The value of tan A is always less than 1 

(a) is false as tan A can be any non-negative value (even greater than 1)

tan 60° = √3 > 1, and tan 90° is undefined (not less than 1). Statements (b), (c), and (d) are all true.

 

Click here to download the free PDF of MCQs worksheet on Chapter 8: Introduction to Trigonometry for Class 10 Maths based on the updated NCERT & CBSE pattern with important multiple-choice questions and answers.

MCQs Worksheet on Chapter 8: Introduction to Trigonometry for Class 10