Heights and Distances: Complete Guide for Competitive Exams

Heights and distances is the direct real-world application of trigonometry. Every concept in this topic — angle of elevation, angle of depression, line of sight — comes from one practical question: given an angle and one measurement, can you find an unknown height or distance?

The answer is always yes, provided you know which trigonometric ratio to apply. And that is the entire skill this topic tests. Unlike pure trigonometry where identities and transformations are required (see our Trigonometry Shortcuts guide), heights and distances uses only three ratios — sin, cos, and tan — applied to right triangles formed by real-world objects.

In SSC CGL Tier 1, heights and distances contributes 1–2 questions. In Tier 2, this increases to 2–3 questions. In NDA, it is a significant topic. In CAT, it appears occasionally within geometry or trigonometry sets. The questions follow fixed templates — once you recognize each template, the solution is a direct substitution.

Part 1: Foundation Concepts

Line of Sight

The line of sight is the straight line from the observer's eye to the object being observed. Every heights and distances problem is built around this line.

Angle of Elevation

When you look up at an object above the horizontal level, the angle between the horizontal line and the line of sight is the angle of elevation.

      Object

         /  |

       /    |

     /      |   height (h)
   /        |
  /         |  
 /  θ      |
--------
Observer   distance (d)

Angle of Depression

When you look down at an object below the horizontal level, the angle between the horizontal line and the line of sight is the angle of depression.

Observer
--------
\   θ    |
 \        |  
  \       |

   \      |   height (h)

     \    |

       \  |
Object    distance (d)   

Key property: Angle of depression from A to B = Angle of elevation from B to A (alternate interior angles).

Standard Trigonometric Values — Must Memorize

Memory trick for sin values: √0/2, √1/2, √2/2, √3/2, √4/2 for 0°, 30°, 45°, 60°, 90°
Cos is sin in reverse order. Tan = sin/cos.

Part 2: The Core Formula — When to Use Which Ratio

Every heights and distances problem gives you one angle and one side of a right triangle. The rule:

In 90% of heights and distances problems — tan is the ratio you need.
Sin and cos appear only when the hypotenuse (direct line of sight distance) is given or required.

Part 3: Single Observer Problems

Type 1 — Find Height Given Distance and Angle of Elevation

Worked Example 1:
A person standing 40m from a tower observes the angle of elevation of the top as 45°. Find the height of the tower.

tan 45° = height / distance
1 = h / 40
h = 40 m

Type 2 — Find Distance Given Height and Angle of Elevation

Worked Example 2:
The angle of elevation of the top of a building 60m tall from a point on the ground is 30°. Find the distance.

tan 30° = 60 / d
1/√3 = 60/d
d = 60√3 = 103.92 m

Type 3 — Angle of Depression from Height

Worked Example 3:
From the top of a cliff 150m high, the angle of depression of a boat is 60°. Find the distance of the boat from the base of the cliff.

Angle of depression = 60° → angle of elevation from boat = 60°
tan 60° = 150 / d
√3 = 150/d
d = 150/√3 = 50√3 = 86.6 m

Type 4 — Finding Hypotenuse (Line of Sight Distance)

Worked Example 4:
A kite is flying at a height of 90m. The string makes an angle of 30° with the ground. Find the length of the string.

sin 30° = height / string length
1/2 = 90 / l
l = 90 × 2 = 180 m

Part 4: Two Position Problems

These are SSC CGL's most tested heights and distances pattern. The observer moves to a new position — and two angles of elevation are given for the same object.

Type 1 — Observer Moves Toward the Tower

Worked Example 5:
A person observes the angle of elevation of a tower as 30°. Walking 40m toward the tower, the angle becomes 60°. Find the height of the tower.

Let height = h, distance from 2nd position = d

From 2nd position: tan 60° = h/d → √3 = h/d → d = h/√3

From 1st position: tan 30° = h/(d+40)
1/√3 = h/(h/√3 + 40)
h/√3 + 40 = h√3
40 = h√3 − h/√3 = h(√3 − 1/√3) = h(3−1)/√3 = 2h/√3
h = 40√3/2 = 20√3 = 34.64 m

Standard shortcut for this type:
h = (d × tan θ₁ × tan θ₂) / (tan θ₂ − tan θ₁)

Where d = distance walked, θ₁ = first angle, θ₂ = second angle

Verify: h = (40 × (1/√3) × √3) / (√3 − 1/√3) = 40 / (2/√3) = 40√3/2 = 20√3 ✓

Type 2 — Two Angles From Same Point (Two Objects)

Worked Example 6:
From a point on the ground, angles of elevation of the top and bottom of a cell tower on a building are 60° and 45° respectively. Building height = 20m. Find tower height.

Let building height = 20m, tower height = t, horizontal distance = d

From angle of bottom of tower (= top of building):
tan 45° = 20/d → 1 = 20/d → d = 20m

From angle of top of tower:
tan 60° = (20+t)/d → √3 = (20+t)/20
20+t = 20√3
t = 20√3 − 20 = 20(√3−1) = 14.64 m

Part 5: Two Observer / Two Tower Problems

Same Base Level — Two Towers Facing Each Other

Worked Example 7:
Two towers A and B stand on the same ground. From the top of tower A (height 30m), angles of depression to the bottom and top of tower B are 60° and 30° respectively. Find height of tower B and distance between them.

Let height of B = h, distance between towers = d

Angle of depression to bottom of B = 60°:
tan 60° = 30/d → √3 = 30/d → d = 30/√3 = 10√3 m

Angle of depression to top of B = 30°:
tan 30° = (30−h)/d → 1/√3 = (30−h)/(10√3)
10√3/√3 = 30−h → 10 = 30−h → h = 20 m

Distance = 10√3 = 17.32 m

Part 6: Shadow Problems

Shadow problems are a specific type where the sun's angle of elevation creates a shadow — the shadow length acts as the horizontal distance.

Worked Example 8:
A vertical pole of height 6m casts a shadow of 6√3 m. Find the angle of elevation of the sun.

tan θ = height/shadow = 6/(6√3) = 1/√3
θ = tan⁻¹(1/√3) = 30°

Worked Example 9:
At a certain time, a 12m pole casts a shadow. Angle of elevation of sun = 45°. Find shadow length.

tan 45° = 12/shadow → 1 = 12/s → s = 12 m

Part 7: Inclined Plane and Slope Problems

Worked Example 10:
A road is inclined at 30° to the horizontal. A person travels 500m along the road. Find the vertical rise.

sin 30° = vertical rise / road length
1/2 = h/500
h = 250 m

Worked Example 11:
From the top of a hill, the angles of depression of two milestones on the road, on opposite sides of the hill, are 30° and 45°. If the hill is 100m high, find the distance between the milestones.

Distance on 30° side: tan 30° = 100/d₁ → d₁ = 100√3 m
Distance on 45° side: tan 45° = 100/d₂ → d₂ = 100 m

Total distance = 100√3 + 100 = 100(√3+1) = 273.2 m

Part 8: Exam Traps and Speed Tips

Trap 1 — Confusing Elevation and Depression

Angle of elevation = observer looks up → object is above observer.
Angle of depression = observer looks down → object is below observer.

The triangle is drawn differently for each — always draw a rough sketch before solving.

Trap 2 — Forgetting Alternate Angles

Angle of depression from A to B equals angle of elevation from B to A. This property is used in almost every two-position problem — not using it leads to incorrect triangle setup.

Trap 3 — Rationalizing the Denominator

Answers like 60/√3 must be rationalized: 60/√3 = 60√3/3 = 20√3. Exam options are always in rationalized form — leaving an answer as 1/√3 will not match any option.

Trap 4 — Same Level Assumption

Heights and distances problems assume the observer's eye level is at ground level unless explicitly stated otherwise. If the observer is standing on a platform or building, add the observer's height to the calculation.

Speed Tip — Standard Angle Results

Memorize these direct results for the most common problem types:

Quick Reference Formula Sheet

Exam-Wise Strategy

2-Week Practice Plan