Theory & Concepts

SSC CGL Heights and Distances Questions, Formulas & Tricks

Get comprehensive theory, expert shortcuts, and hand-picked practice questions for Heights & Distances specifically designed for the SSC CGL 2025-26 pattern.

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35 min readDifficulty: Intermediate

Heights & Distances is the practical application of Trigonometry. Instead of writing long \\( \tan\\theta \\) equations, SSC CGL toppers memorize the standard ratio of sides for 30°, 45°, and 60° triangles to solve these word problems purely through ratios. This module includes 20 massive practice questions with visual diagrams.

Learning path

  • Angle of Elevation vs Depression
  • The 30-60-90 Ratio Trick
  • The 45-45-90 Ratio Trick
  • 20 Full-Length Exam Questions

1. The Golden Ratio Shortcuts

Stop using \\( \tan 30^\\circ = P/B \\). Instead, memorize the ratio of sides for standard right-angled triangles.

The 30° - 60° - 90° Triangle

Sides opposite to angles are in ratio:

1:3:21 : \sqrt{3} : 2

Opposite 30° is 1, Opposite 60° is \\( \sqrt3 \\), Hypotenuse is 2.

The 45° - 45° - 90° Triangle

Sides opposite to angles are in ratio:

1:1:21 : 1 : \sqrt{2}

If angle is 45°, Height = Base!

2. The "Walking Towards" Formula

A classic SSC CGL question: A man walks distance \\( d \\) towards a tower, and the angle of elevation changes from 30° to 60°.

Master Formula

If distance walked is \\( d \\) and height is \\( h \\):

For 30° to 60° shift: d=2h3d = \frac{2h}{\sqrt{3}}

For 45° to 60° shift: d=h(1−13)d = h(1 - \frac{1}{\sqrt{3}})

For 30° to 45° shift: d=h(3−1)d = h(\sqrt{3} - 1)

3. 20 Massive Solved Examples

Question 01Exam Pattern

A tower is 50m high. Its shadow is \( 50\sqrt{3} \) m long. Find the angle of elevation of the sun.

θ50m50√3m
30°
45°
60°
90°
Correct answer: a) 30°

Solution

Step 1: The tangent of the angle of elevation \( \theta \) is \( \frac{\text{Height}}{\text{Shadow}} \).
Step 2: \( \tan\theta = \frac{50}{50\sqrt{3}} = \frac{1}{\sqrt{3}} \).
Final calculation: From standard values, \( \tan 30^\circ = \frac{1}{\sqrt{3}} \). So, \( \theta = 30^\circ \).
Question 02Exam Pattern

A ladder 15m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.

30°WallGround15m
7.5m
7.5√3m
15m
10m
Correct answer: a) 7.5m

Solution

Step 1: The angle is made with the WALL (vertical). This means the angle with the ground is \( 90^\circ - 60^\circ = 30^\circ \).
Step 2: Alternatively, use \( \cos(60^\circ) = \frac{\text{Adjacent}}{\text{Hypotenuse}} \) where Adjacent is the wall.
Step 3: \( \cos 60^\circ = \frac{H}{15} \).
Final calculation: \( \frac{1}{2} = \frac{H}{15} \implies H = 7.5 \)m.
Question 03Exam Pattern

An observer 1.5m tall is 28.5m away from a tower. The angle of elevation from her eyes to the top of the tower is 45°. What is the total height of the tower?

45°h - 1.528.5m
28.5m
30m
31.5m
27m
Correct answer: b) 30m

Solution

Step 1: The angle of elevation is measured from her eyes, which are 1.5m above ground.
Step 2: Base of the top triangle is 28.5m. Angle is 45°.
Step 3: Since \( \tan 45^\circ = 1 \), Perpendicular = Base = 28.5m.
Final calculation: Total Height = Perpendicular + Observer Height = \( 28.5 + 1.5 = 30 \)m.
Question 04Exam Pattern

From a point on the ground, the angle of elevation of the top of a tower is 30°. After walking 20m towards the tower, the angle becomes 60°. Find the height of the tower.

30°60°h20mx
10m
10√3m
20√3m
15m
Correct answer: b) 10√3m

Solution

Step 1: Use the shift formula: Distance walked \( d = 20 \)m.
Step 2: Formula for 30° to 60° shift: \( d = \frac{2h}{\sqrt{3}} \).
Step 3: \( 20 = \frac{2h}{\sqrt{3}} \).
Final calculation: \( 10 = \frac{h}{\sqrt{3}} \implies h = 10\sqrt{3} \)m.
Question 05Exam Pattern

The angles of depression of two ships from the top of a lighthouse are 45° and 30°. If the ships are 200m apart on the same side, find the height of the lighthouse.

30°45°h200mx
100(√3 - 1)m
100(√3 + 1)m
200(√3 - 1)m
200(√3 + 1)m
Correct answer: b) 100(√3 + 1)m

Solution

Step 1: Inner triangle is 45°, so Base = Height = \( h \).
Step 2: Outer triangle is 30°, so Total Base = \( h\sqrt{3} \).
Step 3: Distance between ships is \( \text{Total Base} - \text{Inner Base} = h\sqrt{3} - h = h(\sqrt{3} - 1) \).
Step 4: \( h(\sqrt{3} - 1) = 200 \).
Final calculation: \( h = \frac{200}{\sqrt{3} - 1} = \frac{200(\sqrt{3} + 1)}{2} = 100(\sqrt{3} + 1) \)m.
Question 06Exam Pattern

A kite is flying at a height of 60m. The string makes an angle of 60° with the ground. Find the length of the string.

60°60mL
40√3m
30√3m
20√3m
60√3m
Correct answer: a) 40√3m

Solution

Step 1: We know Perpendicular = 60m and want the Hypotenuse (String).
Step 2: Use sine: \( \sin 60^\circ = \frac{P}{H} \).
Step 3: \( \frac{\sqrt{3}}{2} = \frac{60}{L} \).
Final calculation: \( L = \frac{120}{\sqrt{3}} = 40\sqrt{3} \)m.
Question 07Exam Pattern

The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 4.6m away from the wall. The length of the ladder is:

60°4.6mL
2.3m
4.6m
9.2m
13.8m
Correct answer: c) 9.2m

Solution

Step 1: Base = 4.6m. We want the Hypotenuse (Ladder).
Step 2: Use cosine: \( \cos 60^\circ = \frac{B}{H} \).
Step 3: \( \frac{1}{2} = \frac{4.6}{L} \).
Final calculation: \( L = 4.6 \times 2 = 9.2 \)m.
Question 08Exam Pattern

If the height of a pole is \( 2\sqrt{3} \) meters and the length of its shadow is 2 meters, find the angle of elevation.

30°
45°
60°
90°
Correct answer: c) 60°

Solution

Step 1: \( \tan\theta = \frac{P}{B} \).
Step 2: \( P = 2\sqrt{3} \) and \( B = 2 \).
Step 3: \( \tan\theta = \frac{2\sqrt{3}}{2} = \sqrt{3} \).
Final calculation: \( \tan 60^\circ = \sqrt{3} \), so \( \theta = 60^\circ \).
Question 09Exam Pattern

The shadow of a tower is \( \sqrt{3} \) times its height. The angle of elevation of the sun is:

30°
45°
60°
90°
Correct answer: a) 30°

Solution

Step 1: Let height = \( h \). Then shadow = \( h\sqrt{3} \).
Step 2: \( \tan\theta = \frac{\text{Height}}{\text{Shadow}} = \frac{h}{h\sqrt{3}} = \frac{1}{\sqrt{3}} \).
Final calculation: \( \tan 30^\circ = \frac{1}{\sqrt{3}} \), so angle is 30°.
Question 10Exam Pattern

From the top of a 7m high building, the angle of elevation of the top of a tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

7(√3 + 1)m
7(√3 - 1)m
7√3m
14m
Correct answer: a) 7(√3 + 1)m

Solution

Step 1: The angle of depression to the foot is 45°, which means the horizontal distance (Base) equals the height of the building (7m).
Step 2: Now look at the triangle above the building. Base is 7m, angle of elevation is 60°.
Step 3: Height of this top part = \( 7 \times \tan 60^\circ = 7\sqrt{3} \)m.
Final calculation: Total height = Top part + Bottom part = \( 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \)m.
Question 11Exam Pattern

Two poles of equal heights are standing opposite each other on either side of a road, which is 80m wide. From a point between them, angles of elevation are 60° and 30°. Find the height of the poles.

10√3m
20√3m
40√3m
30√3m
Correct answer: b) 20√3m

Solution

Step 1: Let the distances from the point be \( x \) and \( 80-x \). Heights are \( h \).
Step 2: \( \tan 60^\circ = h/x \implies h = x\sqrt{3} \).
Step 3: \( \tan 30^\circ = h/(80-x) \implies h = (80-x)/\sqrt{3} \).
Step 4: Equate them: \( x\sqrt{3} = (80-x)/\sqrt{3} \implies 3x = 80 - x \implies 4x = 80 \implies x = 20 \).
Final calculation: \( h = 20\sqrt{3} \)m.
Question 12Exam Pattern

A tree breaks due to a storm and bends so the top touches the ground at a 30° angle. The foot to the touch point is 8m. Find total height.

8√3m
16√3m
24√3m
8m
Correct answer: a) 8√3m

Solution

Step 1: Base = 8m. Angle = 30°.
Step 2: Standing part (Perpendicular) = \( 8 \times \tan 30^\circ = 8/\sqrt{3} \).
Step 3: Broken part (Hypotenuse) = \( 8 / \cos 30^\circ = 8 / (\sqrt{3}/2) = 16/\sqrt{3} \).
Final calculation: Total height = \( 8/\sqrt{3} + 16/\sqrt{3} = 24/\sqrt{3} = 8\sqrt{3} \)m.
Question 13Exam Pattern

From a 120m high tower, a man observes two cars on opposite sides with angles of depression 60° and 45°. Find distance between them.

120(1 + 1/√3)m
120(1 - 1/√3)m
120√3m
120m
Correct answer: a) 120(1 + 1/√3)m

Solution

Step 1: This is the 'opposite sides' scenario. Add the bases.
Step 2: Base 1 (45° side) = Height = 120m.
Step 3: Base 2 (60° side) = \( \frac{120}{\tan 60^\circ} = \frac{120}{\sqrt{3}} \).
Final calculation: Total Distance = \( 120 + \frac{120}{\sqrt{3}} = 120(1 + \frac{1}{\sqrt{3}}) \)m.
Question 14Exam Pattern

An aeroplane at 3000m is vertically above another. Angles of elevation from ground are 60° and 45°. Find distance between planes.

3000(1 - 1/√3)m
1000(3 - √3)m
1000(3 + √3)m
3000m
Correct answer: b) 1000(3 - √3)m

Solution

Step 1: The top plane is at 3000m (angle 60°). So the Base on ground = \( 3000 / \sqrt{3} = 1000\sqrt{3} \)m.
Step 2: The bottom plane has angle 45°. Its height = Base = \( 1000\sqrt{3} \)m.
Final calculation: Distance between them = \( 3000 - 1000\sqrt{3} = 1000(3 - \sqrt{3}) \)m.
Question 15Exam Pattern

Angle of elevation of a tower from points 4m and 9m away in same line are complementary. Find height.

5m
6m
13m
36m
Correct answer: b) 6m

Solution

Step 1: Standard SSC trick: If distances are \( a \) and \( b \) and angles are complementary (sum to 90°).
Step 2: The height \( h \) is given perfectly by \( h = \sqrt{ab} \).
Final calculation: \( h = \sqrt{4 \times 9} = \sqrt{36} = 6 \)m.
Question 16Exam Pattern

Highway leads to 50m tower. From top, angles of depression of two cars are 30° and 60°. Distance between cars?

100/√3 m
50/√3 m
200/√3 m
150/√3 m
Correct answer: a) 100/√3 m

Solution

Step 1: Base for 30° car = \( 50\sqrt{3} \)m.
Step 2: Base for 60° car = \( 50/\sqrt{3} \)m.
Step 3: They are on the same highway (same side). Distance = difference.
Final calculation: \( 50\sqrt{3} - \frac{50}{\sqrt{3}} = \frac{150 - 50}{\sqrt{3}} = \frac{100}{\sqrt{3}} \)m.
Question 17Exam Pattern

A cloud is at elevation 30° from 60m above lake. Reflection depression is 60°. Find height of cloud from lake.

60m
120m
180m
240m
Correct answer: b) 120m

Solution

Step 1: Master formula: \( H = h \times \frac{\tan\beta + \tan\alpha}{\tan\beta - \tan\alpha} \).
Step 2: \( \alpha = 30^\circ \), \( \beta = 60^\circ \), \( h = 60 \).
Step 3: \( \tan 60^\circ = \sqrt{3} \), \( \tan 30^\circ = 1/\sqrt{3} \).
Step 4: \( H = 60 \times \frac{\sqrt{3} + 1/\sqrt{3}}{\sqrt{3} - 1/\sqrt{3}} = 60 \times \frac{4/\sqrt{3}}{2/\sqrt{3}} = 60 \times 2 \).
Final calculation: 120m.
Question 18Exam Pattern

A flagstaff 5m high stands on a building. Angles of elevation of top and bottom are 60° and 45°. Height of building?

5(√3 - 1)/2 m
5(√3 + 1)/2 m
5m
2.5m
Correct answer: b) 5(√3 + 1)/2 m

Solution

Step 1: Let building height = \( x \). Base = \( x \) (because 45° angle).
Step 2: Total height = \( x + 5 \). Angle is 60°.
Step 3: \( \tan 60^\circ = \frac{x + 5}{x} \implies \sqrt{3}x = x + 5 \).
Step 4: \( x(\sqrt{3} - 1) = 5 \).
Final calculation: \( x = \frac{5}{\sqrt{3}-1} = \frac{5(\sqrt{3}+1)}{2} \)m.
Question 19Exam Pattern

Two ships on opposite sides of a 100m lighthouse have angles of elevation 30° and 45°. Distance between them?

100(√3 - 1)m
100(√3 + 1)m
200m
150m
Correct answer: b) 100(√3 + 1)m

Solution

Step 1: Base on 45° side = Height = 100m.
Step 2: Base on 30° side = \( 100\sqrt{3} \)m.
Step 3: Opposite sides mean we add the bases.
Final calculation: Distance = \( 100 + 100\sqrt{3} = 100(\sqrt{3} + 1) \)m.
Question 20Exam Pattern

A bird at height \( 50\sqrt{3} \)m is observed in North at 30° and South at 60° after 2 min. Speed in m/min?

50
100
150
200
Correct answer: b) 100

Solution

Step 1: Bird crosses from North to South (opposite sides of observer).
Step 2: Distance North = \( \frac{50\sqrt{3}}{\tan 30^\circ} = 50\sqrt{3} \times \sqrt{3} = 150 \)m.
Step 3: Distance South = \( \frac{50\sqrt{3}}{\tan 60^\circ} = \frac{50\sqrt{3}}{\sqrt{3}} = 50 \)m.
Step 4: Total distance = 150 + 50 = 200m.
Final calculation: Speed = \( \frac{200\text{m}}{2\text{min}} = 100 \) m/min.