Theory & Concepts

SSC CGL Angles of Elevation and Depression Questions & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Angles of Elevation/Depression specifically designed for the SSC CGL 2025-26 pattern.

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25 min readDifficulty: Intermediate

Mastering the Angle of Elevation and Depression is about more than just theory. It's about translating a word problem into a precise geometric diagram. In this module, we move from the core definitions into 10 high-impact exam questions, featuring visual diagrams for 7 Foundation-level scenarios.

Learning path

  • Line of Sight Mapping
  • Elevation to Depression Translation
  • 7 Diagram-Based Practice Probs
  • 10 Total Solved Examples

Practice Questions with Diagrams

Question 01Exam Pattern

A tower is 75m high. If the angle of elevation of the sun is 60°, find the length of the shadow.

60°75mShadow
25√3m
75√3m
25m
50m
Correct answer: a) 25√3m

Solution

Step 1: In the right triangle, \( \tan 60^\circ = \frac{\text{Height}}{\text{Shadow}} \).
Step 2: \( \sqrt{3} = \frac{75}{x} \).
Final calculation: \( x = \frac{75}{\sqrt{3}} = 25\sqrt{3} \)m.
Question 02Exam Pattern

A kite is flying at a height of 120m. The string attached to it makes an angle of 30° with the ground. Find the length of the string.

30°120mString
240m
120√3m
60√3m
240√3m
Correct answer: a) 240m

Solution

Step 1: We have Perpendicular (120m) and need Hypotenuse (String).
Step 2: Use sine: \( \sin 30^\circ = \frac{P}{H} \).
Step 3: \( \frac{1}{2} = \frac{120}{L} \).
Final calculation: \( L = 120 \times 2 = 240 \)m.
Question 03Exam Pattern

A person 1.6m tall stands 20m away from a wall. The angle of elevation from their eyes to the top of the wall is 45°. Find the height of the wall.

45°h - 1.620m
20m
21.6m
18.4m
23.2m
Correct answer: b) 21.6m

Solution

Step 1: For the triangle above eye level, \( \tan 45^\circ = 1 \), so Perpendicular = Base = 20m.
Step 2: Total height = Triangle Height + Person's Height.
Final calculation: \( 20 + 1.6 = 21.6 \)m.
Question 04Exam Pattern

The angle of elevation of the top of a pillar changes from 30° to 45° as an observer walks 50m towards it. Find the height of the pillar.

30°45°h50mh
25(√3+1)m
25(√3-1)m
50(√3+1)m
50(√3-1)m
Correct answer: a) 25(√3+1)m

Solution

Step 1: Distance \( d = h(\cot 30^\circ - \cot 45^\circ) \).
Step 2: \( 50 = h(\sqrt{3} - 1) \).
Step 3: \( h = \frac{50}{\sqrt{3}-1} = \frac{50(\sqrt{3}+1)}{2} \).
Final calculation: \( h = 25(\sqrt{3} + 1) \)m.
Question 05Exam Pattern

From the top of a 90m high balloon, the angle of depression of two cars on the ground is 60° and 30°. Find the distance between the cars.

30°60°90mdx
60√3m
30√3m
90√3m
45√3m
Correct answer: a) 60√3m

Solution

Step 1: Base of 60° car = \( 90/\sqrt{3} = 30\sqrt{3} \)m.
Step 2: Base of 30° car = \( 90\sqrt{3} \)m.
Final calculation: Distance = \( 90\sqrt{3} - 30\sqrt{3} = 60\sqrt{3} \)m.
Question 06Exam Pattern

From the top of a 10m high pillar, the angle of elevation of a tower top is 60° and depression of its foot is 30°. Find tower height.

30°60°Top Part10mBase
30m
40m
20m
25m
Correct answer: b) 40m

Solution

Step 1: The 30° depression to the foot means \( \tan 30^\circ = 10/\text{Base} \implies \text{Base} = 10\sqrt{3} \)m.
Step 2: Top Part of tower = \( \text{Base} \times \tan 60^\circ = 10\sqrt{3} \times \sqrt{3} = 30 \)m.
Final calculation: Total height = Top part + Bottom part (10m) = \( 30 + 10 = 40 \)m.
Question 07Exam Pattern

The angle of elevation of the top of a tower from a point A is 45°. Walking 10m away from the tower, it becomes 30°. Height?

30°45°h10mh
5(√3+1)m
5(√3-1)m
10(√3+1)m
10(√3-1)m
Correct answer: a) 5(√3+1)m

Solution

Step 1: \( 10 = h(\cot 30^\circ - \cot 45^\circ) = h(\sqrt{3}-1) \).
Final calculation: \( h = \frac{10}{\sqrt{3}-1} = 5(\sqrt{3}+1) \)m.
Question 08Exam Pattern

If the angle of depression of a boat from the top of a 60m lighthouse is 30°, what is the distance of the boat from the foot of the lighthouse?

60m
60√3m
20√3m
120m
Correct answer: b) 60√3m

Solution

Step 1: Angle of Depression (30°) = Angle of Elevation from boat.
Step 2: \( \tan 30^\circ = \frac{60}{x} \implies \frac{1}{\sqrt{3}} = \frac{60}{x} \).
Final calculation: \( x = 60\sqrt{3} \)m.
Question 09Exam Pattern

The angles of elevation of a tower from two points at distances of 9m and 16m are complementary. Find height.

12m
144m
25m
7m
Correct answer: a) 12m

Solution

Step 1: Use the standard trick for complementary angles: \( h = \sqrt{ab} \).
Final calculation: \( h = \sqrt{9 \times 16} = 12 \)m.
Question 10Exam Pattern

A car moving at constant speed towards a tower takes 12 mins for angle of elevation to change from 30° to 60°. How much more time to reach the tower?

6 mins
12 mins
4 mins
8 mins
Correct answer: a) 6 mins

Solution

Step 1: Let the speed be \( v \) and height be \( h \). Distance 30° to 60° is \( 2h/\sqrt{3} \).
Step 2: Time taken is 12 mins. Remaining distance is \( h/\sqrt{3} \).
Step 3: Notice that remaining distance is exactly HALF of the distance already covered.
Final calculation: Time taken will also be half: \( 12/2 = 6 \) mins.