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SSC CGL Heights & Distances

Study Material — 4 PYQs (2018–2020) · Concept Notes · Shortcuts

SSC CGL Heights & Distances is a frequently tested subtopic — 4 previous year questions from 2018–2020 papers are included below with concept notes, key rules and shortcut tricks.

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Previous Year Questions

SSC CGL Heights & Distances — Past Exam Questions

4 questions from actual SSC CGL papers · all shown free · click option to reveal solution

Exam Q 12020Previous Year Pattern

A man standing 30 metres away from the base of a tower observes the angle of elevation to the top of the tower to be 60°. Find the height of the tower. (Use √3 ≈ 1.732)

Exam Q 22019Previous Year Pattern

A man standing 40 metres away from the base of a tower observes that the angle of elevation to the top of the tower is 30°. Find the height of the tower. (Use √3 ≈ 1.732)

Exam Q 32018Previous Year Pattern

From the top of a lighthouse 60 m high, the angles of depression of two ships on the same side are 45° and 30°. What is the distance between the two ships?

Exam Q 42019Previous Year Pattern

From a point on the ground 40 m away from the base of a vertical tower, the angle of elevation to the top is 60°. A man climbs to a point on the tower such that the angle of elevation from the same ground point becomes 45°. How many metres did the man climb?

Concept Notes

Heights & Distances— Rules & Concept

Core ConceptRead this first — the foundation of the topic

HEIGHTS AND DISTANCES — COMPLETE GUIDE FOR SSC CGL ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

CORE CONCEPT ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Heights and Distances is a real-life application of Trigonometry. We use it to find the height of a building, tower, tree, or mountain — without physically measuring it. We also find the distance between two objects. The tool we use is the angle formed between the line of sight and the horizontal ground. Think of it this way: You stand on the ground and look up at the top of a tower. The angle your eyes make with the horizontal is the Angle of Elevation. Now imagine you are standing on the top of a building and looking DOWN at a car on the road. That angle is the Angle of Depression.

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ KEY DEFINITIONS

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ • Angle of Elevation: The angle formed when you look UPWARD from the horizontal. It is measured from the horizontal line UP to the line of sight.

• Angle of Depression: The angle formed when you look DOWNWARD from the horizontal. It is measured from the horizontal line DOWN to the line of sight. • Line of Sight: The straight line from your eye to the object.

• Horizontal Line: The flat line parallel to the ground from your eye level.

Key RulesCore rules you must know cold

The Angle of Elevation from point A to point B equals the Angle of Depression from point B to point A. They are alternate interior angles and are always EQUAL. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Formula BlockMemorise — at least one formula appears in every paper

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

In a right triangle: tan(angle) = Opposite / Adjacent = Height / Distance

Key Trigonometric Values (MEMORISE THESE):

• tan 30° = 1/√3 ≈ 0.577
• tan 45° = 1
• tan 60° = √3 ≈ 1.732
• sin 30° = 1/2, cos 30° = √3/2
• sin 45° = 1/√2, cos 45° = 1/√2
• sin 60° = √3/2, cos 60° = 1/2
Core formula: Height = Distance × tan(angle of elevation)
Core formula: Distance = Height / tan(angle of elevation)

━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Exam PatternsWhat examiners ask — read before attempting PYQs

— WHAT SSC ASKS ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SSC CGL asks 1–2 questions from this topic every year. The most common patterns are: 1. One observer, one tower — find height or distance. 2. Two angles of elevation from two different points — find height. 3.

Tower on top of a hill or building — combined height problems. 4. Two observers on opposite sides of a tower. 5. Shadow-based problems using sun's angle of elevation. ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SHORTCUTS AND TRICKS ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ SHORTCUT 1 — Two Angle Formula (Most Tested): When a tower of height H is observed from two points at distances d1 and d2 on the same side, with angles A and B: H² = d1 × d2 × tan(A) × tan(B) Special case: If angles are complementary (A + B = 90°), then H = √(d1 × d2) SHORTCUT 2 — 45° Shortcut: When angle of elevation = 45°, tan 45° = 1, so Height = Distance. This is the fastest case.

No calculation needed. SHORTCUT 3 — Shadow Length Trick: When the sun makes angle θ with the ground: Height of object = Shadow length × tan(θ) If θ = 60°, Height = Shadow × √3 If θ = 30°, Height = Shadow / √3 ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Worked ExampleSolve this step-by-step before moving on
1
Step 1

Draw the situation. You have a right triangle. The base (horizontal distance) = 40 m. The angle at the base = 30°. Height = H (unknown).

2
Step 2

Use formula — tan(angle) = Height / Distance tan 30° = H / 40 1/√3 = H / 40 H = 40 / √3 H = 40√3 / 3 H = 40 × 1.732 / 3 ≈ 23.09 m Final Answer: Height of tower = 40/√3 = 40√3/3 meters ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ WORKED EXAMPLE 2 — Two Angle Problem ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Question: The angle of elevation of the top of a tower from point A is 45°. From point B, which is 20 m closer to the tower, the angle of elevation is 60°. Find the height of the tower.

1
Step 1

Let height = H. Let distance from B to base of tower = d. From B: tan 60° = H/d → √3 = H/d → d = H/√3 From A: tan 45° = H/(d + 20) → 1 = H/(d + 20) → d + 20 = H

2
Step 2

Substitute d = H/√3 into second equation. H/√3 + 20 = H 20 = H − H/√3 20 = H(1 − 1/√3) 20 = H × (√3 − 1)/√3 H = 20√3 / (√3 − 1)

3
Step 3

Rationalise — Multiply numerator and denominator by (√3 + 1). H = 20√3(√3 + 1) / [(√3 − 1)(√3 + 1)] H = 20√3(√3 + 1) / (3 − 1) H = 20√3(√3 + 1) / 2 H = 10√3(√3 + 1) H = 10(3 + √3) H = 30 + 10√3 meters Final Answer: Height = 10(3 + √3) = approximately 47.32 m ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

Exam TrapsCommon mistakes students make — avoid these

— THE NUMBER 1 TRAP ━━━━━━━━━━━━━━━━━━━━━━━━━━━━━ Students confuse Angle of Depression with Angle of Elevation in two-observer problems. They draw the angle of depression from the wrong position or forget that the angle of depression from the top equals the angle of elevation from the bottom. Always draw the figure first.

Never solve without a diagram. Many errors are simply diagram errors, not calculation errors.

Key Points to Remember

  • Angle of Elevation = angle formed when looking UPWARD from horizontal; Angle of Depression = angle formed when looking DOWNWARD from horizontal.
  • Angle of Elevation from point A to B = Angle of Depression from point B to A (alternate interior angles — always equal).
  • Formula: tan(angle) = Height / Distance. This is the master formula for all basic problems.
  • Shortcut: When angle = 45°, tan 45° = 1, so Height = Distance directly — no calculation needed.
  • Shortcut: When two angles are complementary (add up to 90°) and tower height H is observed from distances d1 and d2, then H = √(d1 × d2).
  • Key values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. These three cover 90% of all exam questions.
  • Shadow problem formula: Height = Shadow length × tan(sun's angle of elevation).
  • Always RATIONALISE the denominator when answer contains expressions like 1/(√3 − 1) — SSC answer options are always in rationalised form.
  • In two-angle problems, set up two equations using tan, then solve simultaneously by substitution.
  • Draw the figure FIRST before writing any equation — most errors in this topic come from wrong diagrams, not wrong formulas.

Exam-Specific Tips

  • tan 30° = 1/√3 = √3/3 ≈ 0.5774; this is used in the most frequently appearing tower-distance problems in SSC CGL.
  • When angle of elevation of the sun is 60°, the length of the shadow of a pole of height H is H/√3 (i.e., shadow = Height/√3).
  • The angle of elevation and angle of depression between two mutually observable points are always equal — this is a direct consequence of alternate interior angles with a transversal cutting parallel lines.
  • For a tower of height H observed from two points on the same horizontal line at distances d1 and d2 with angles of elevation cot A and cot B respectively, the formula is: H²(cot²A − cot²B) = d² where d is the distance between the two points.
  • When a ladder of length L leans against a wall making angle θ with the ground: height reached = L × sin θ and distance from wall = L × cos θ.
  • If from the top of a cliff of height h, angles of depression of the top and bottom of a tower are α and β respectively, then tower height = h × (tan β − tan α) / tan β.
  • sin 45° = cos 45° = 1/√2 ≈ 0.7071; these values appear in problems where the observer and object are at equal heights or distances.
Practice MCQs

Heights & Distances — Practice Questions

45graded MCQs · easy to hard · full solution & trap analysis · showing 20 of 45

All MCQs →
Practice 1easy

From the top of a cliff 80 metres high, the angle of depression to a boat on the water is 30°. How far is the boat from the base of the cliff?

Practice 2easy

A ladder leans against a wall such that it makes an angle of 45° with the ground. If the ladder is 10√2 metres long, what is the height at which the ladder touches the wall?

Practice 3easy

From a point on the ground 50 metres away from the base of a building, the angle of elevation to the top is 45°. What is the height of the building?

Practice 4easy

An observer on the ground sees the top of a tree at an angle of elevation of 30°. If the observer moves 20 metres closer to the tree, the angle of elevation becomes 60°. Find the height of the tree.

Practice 5easy

A person standing on the roof of a 20-metre-high building observes the angle of depression to a point on the ground to be 45°. What is the horizontal distance from the building to that point?

Practice 6easy

From the top of a 40-metre-high cliff, a person observes a boat in the sea at an angle of depression of 30°. How far is the boat from the base of the cliff?

Practice 7easy

A ladder leans against a wall such that its foot is 6 metres away from the wall. The angle of elevation from the foot of the ladder to the top of the wall is 45°. What is the height of the wall?

Practice 8easy

A vertical pole of height 15 metres casts a shadow of 15 metres on the ground. What is the angle of elevation of the sun?

Practice 9easy

An observer standing on the ground sees the top of a tree at an angle of elevation of 30°. If the observer moves 10 metres closer to the tree, the angle of elevation becomes 60°. Find the height of the tree.

Practice 10easy

From a point on the ground, the angle of elevation to the top of a 20-metre tall building is 45°. How far is the point from the base of the building?

Practice 11easy

A man standing 30 metres away from the base of a tower observes the angle of elevation to the top of the tower to be 60°. Find the height of the tower.

Practice 12easy

From the top of a cliff 80 metres high, the angle of depression to a boat in the sea is 30°. How far is the boat from the base of the cliff? (Use √3 = 1.732)

Practice 13easy

A man standing 30 metres away from the base of a tower observes that the angle of elevation to the top of the tower is 60°. Find the height of the tower. (Use √3 = 1.732)

Practice 14easy

A man standing 30 metres away from the base of a tower observes the angle of elevation to the top of the tower to be 60°. Find the height of the tower.

Practice 15easy

From the top of a cliff 80 metres high, the angle of depression to a boat in the sea is 30°. What is the horizontal distance of the boat from the base of the cliff?

Practice 16easy

A man standing 30 metres away from the base of a tower observes the angle of elevation to the top of the tower as 60°. Find the height of the tower.

Practice 17easy

Two buildings are 50 metres apart. From the top of the first building (height 30 m), the angle of depression to the top of the second building is 15°. What is the height of the second building? (Use tan(15°) ≈ 0.27)

Practice 18medium

Two buildings stand on level ground. From the top of the first building (height 20 m), the angle of elevation to the top of the second building is 30°. The horizontal distance between the buildings is 20√3 metres. What is the height of the second building?

Practice 19medium

A ladder leans against a wall. The angle between the ladder and the ground is 60°. If the ladder is 10 m long, how high up the wall does the ladder reach?

Practice 20medium

From a point on the ground, the angle of elevation to the top of a building is 45°. From a point 30 m further away (on the same line), the angle of elevation is 30°. What is the height of the building?

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60-Second Revision — Heights & Distances

  • Formula: tan(angle) = Height / Distance — this single formula solves 70% of all Heights and Distances questions.
  • Remember: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. Write these on your rough sheet at the start of the exam.
  • Shortcut: If angle = 45°, Height = Distance instantly. If angles are complementary, Height = √(d1 × d2).
  • Trap: Angle of Depression is measured from the HORIZONTAL downward — NOT from the vertical. Never add it to 90°. Draw the figure first.
  • Remember: Angle of Elevation (from below) = Angle of Depression (from above) for the same two points — use this to avoid confusion in two-building problems.
  • Always rationalise your final answer — SSC options never have surds in the denominator.
  • In two-angle problems: write two tan equations, substitute one into the other, and solve for H. This method works for every such problem without exception.
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