Trigonometry and Heights & Distances Quiz: SSC Practice
10 timed trigonometry questions for SSC CGL and CHSL. Standard values, identities, complementary angles, maximum values, and heights and distances.

Trigonometry and Heights & Distances Quiz: SSC Practice
Tip: use keys 1-4 to answer, Enter for next.
Answers to the starter quiz
Spoilers! These are the Starter quiz’s answers. Easy, Medium, Hard and Mixed rounds show their own answers when you finish.
What is the value of sin²60° + cos²45°?
- 1
- 5/4
- 3/2
- 7/4
Answer: 5/4. sin 60° = sqrt(3)/2 so sin²60° = 3/4; cos 45° = 1/sqrt(2) so cos²45° = 1/2. Total 5/4.
If sin A = 3/5 and A is acute, what is tan A?
- 3/4
- 4/3
- 4/5
- 5/3
Answer: 3/4. Opposite 3, hypotenuse 5, so adjacent = 4 (3-4-5 triangle). tan A = 3/4.
If sec A + tan A = 3, what is the value of sec A − tan A?
- 3
- 2/3
- 1/3
- 1/9
Answer: 1/3. (sec A + tan A)(sec A − tan A) = sec²A − tan²A = 1, so sec A − tan A = 1/3.
If sin 3A = cos (A − 10°), where 3A is acute, what is A?
- 20°
- 25°
- 30°
- 22.5°
Answer: 25°. sin x = cos y with acute angles means x + y = 90°: 3A + A − 10° = 90°, so A = 25°.
What is the value of tan 1° × tan 2° × tan 3° × ... × tan 89°?
- 0
- 89
- 1/2
- 1
Answer: 1. Pair tan A with tan (90° − A) = cot A; each pair multiplies to 1, and the middle term tan 45° = 1.
When the sun's elevation is 60°, a tower casts a shadow 20 m long. What is the height of the tower?
- 20 sqrt(3) m
- 20/sqrt(3) m
- 40 m
- 10 sqrt(3) m
Answer: 20 sqrt(3) m. tan 60° = height/shadow, so height = 20 × sqrt(3) = 20 sqrt(3) m (about 34.6 m).
A 10 m ladder leans against a wall and makes an angle of 60° with the ground. How high up the wall does it reach?
- 5 m
- 10 sqrt(3) m
- 5 sqrt(3) m
- 5 sqrt(2) m
Answer: 5 sqrt(3) m. Height = ladder × sin 60° = 10 × sqrt(3)/2 = 5 sqrt(3) m.
The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 60 m towards the tower, it becomes 60°. What is the height of the tower?
- 20 sqrt(3) m
- 60 m
- 30 m
- 30 sqrt(3) m
Answer: 30 sqrt(3) m. h cot 30° − h cot 60° = 60 gives h(sqrt(3) − 1/sqrt(3)) = 60, so h × 2/sqrt(3) = 60 and h = 30 sqrt(3) m.
What is the maximum value of 5 sin A + 12 cos A?
- 17
- 12
- 13
- 7
Answer: 13. a sin A + b cos A has maximum sqrt(a² + b²) = sqrt(25 + 144) = 13.
From the top of a cliff 50 m high, the angle of depression of a boat is 30°. How far is the boat from the foot of the cliff?
- 50 m
- 50 sqrt(3) m
- 50/sqrt(3) m
- 100 m
Answer: 50 sqrt(3) m. The angle of depression equals the angle of elevation from the boat. tan 30° = 50/d, so d = 50 sqrt(3) m (about 86.6 m).
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