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CBSE · Class 10 · Mathematics

Introduction to Trigonometry

Standard difficulty · 20 questions · 40 marks

Name: Date: Score: ____ / 40

Show your working for Sections B and C. Answers are on the last page — try every question first!

Section A — Choose the correct answer8 × 1 = 8 marks
1.
The value of sin 45° is:
  1. 1/√2
  2. 0
  3. √3
  4. 1/√3
2.
In △ABC, right-angled at B, AB = 180 cm, BC = 33 cm and AC = 183 cm. sin A equals:
  1. 60/11
  2. 60/61
  3. 61/11
  4. 11/61
3.
If tan A = 5/12, then cos A equals:
  1. 13/12
  2. 13/5
  3. 12/13
  4. 12/5
4.
If cosec A = 29/20, then tan A equals:
  1. 21/20
  2. 20/21
  3. 20/29
  4. 21/29
5.
sin2 A + cos2 A equals:
  1. tan2 A
  2. 0
  3. 1
  4. 2
6.
The value of 2 cos2 60° − 3 cos2 0° is:
  1. −3/2
  2. 7/2
  3. −5/2
  4. −5
7.
The value of tan 60° cos 30° + cos 60° tan 45° is:
  1. 2
  2. 1
  3. 3/2
  4. 4
8.
Assertion (A): If sin A = 35/37, then cos A = 35/12.
Reason (R): For every acute angle A, sin2 A + cos2 A = 1.
  1. Both A and R are true, and R is the correct explanation of A.
  2. Both A and R are true, but R is not the correct explanation of A.
  3. A is true, but R is false.
  4. A is false, but R is true.
Section B — Short answer8 × 2 = 16 marks
9.
If sin A = 4/5, find cos A and tan A.
10.
Evaluate: (sin2 0° + 2 cos2 60°) ÷ (cos2 30°)
11.
If tan A = 60/11, find sin A and cos A.
12.
If 2 tan A = 2√3, where A is acute, find A.
13.
If cos(A − B) = 1/2 and cos(A + B) = 0, with 0° < A + B ≤ 90° and A > B, find A and B.
14.
Evaluate: (cos 60° − sin 60°)2
15.
In △ABC, right-angled at B, AC = 18 cm and ∠A = 45°. Find BC and AB.
16.
If cot θ = 20/21, evaluate ((1 + sin θ)(1 − sin θ)) ÷ ((1 + cos θ)(1 − cos θ)).
Section C — Word problems4 × 4 = 16 marks
17.
From a point on the ground 15 m from the foot of a tower, the angle of elevation of its top is 30°. Find the height of the tower.
18.
A kite flies on a string 122 m long, which makes an angle of 30° with the ground. Assuming the string is straight, how high is the kite?
19.
A ladder leans against a wall and makes an angle of 60° with the ground. Its foot is 5 m from the wall. How long is the ladder?
20.
From the top of a building 60 m high, the angle of depression of a car on the road is 60°. How far is the car from the foot of the building?
AmbifulWorksheets
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CBSE · Class 10 · Mathematics

Answer key

Introduction to Trigonometry · AW-10M-TRI-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.

QAnswerWorking
1(a) 1/√2standard value: sin 45° = 1/√2
2(d) 11/61sin A = BC/AC = 11/61
3(c) 12/13sides in the ratio 5 : 12 : 13 (opposite : adjacent : hypotenuse)
4(b) 20/21cosec A = 29/20 gives sides 20 : 21 : 29; tan A = 20/21
5(c) 1for every acute angle A
6(c) −5/22 cos2 60° = 1/2, 3 cos2 0° = 3
7(a) 23/2 + 1/2
8(d) A is false, but R is true.A is false: cos A = √(1 − sin2 A) = 12/37, not 35/12. R is true.
9cos A = 3/5, tan A = 4/3adjacent = √(52 − 42) = 3
102/31/2 ÷ 3/4 = 2/3
11sin A = 60/61, cos A = 11/61hypotenuse = √(602 + 112) = 61
12A = 60°tan A = √3, so A = 60°
13A = 75°, B = 15°A − B = 60°, A + B = 90°
141 − √3/2sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3
15BC = 9√2 cm, AB = 9√2 cmBC = AC sin A, AB = AC cos A
16400/441= (1 − sin2 θ)/(1 − cos2 θ) = cos2 θ/sin2 θ = cot2 θ = 400/441
175√3 mheight = 15 × tan 30° = 15 × 1/√3
1861 mheight = 122 × sin 30°
1910 mladder = 5 ÷ cos 60° = 5 ÷ (1/2)
2020√3 mdistance = 60 ÷ tan 60°