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AFCAT Heights & Distances

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This page covers AFCAT Heights & Distances with complete concept notes, 28 graded practice MCQs, key points and exam-specific tips. Free to study.

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Concept Notes

Heights & Distances— Rules & Concept

1.Core Concept

Read this first — the foundation of the topic
Example

looking up at the top of a tower. • Angle of Depression: The angle formed when you look DOWN at an object from a horizontal line

Example

a person on a cliff looking down at a boat. • Line of Sight: The straight line from your eye to the object. • Horizontal Line: The flat line from your eye level going straight ahead.

2.Key Rules

Core rules you must know cold

Angle of elevation from point A to point B = Angle of depression from point B to point A. They are always equal (alternate interior angles).

3.Formula Block

Memorise — at least one formula appears in every paper
• tan(angle) = Opposite / Adjacent = Height / Base Distance
• sin(angle) = Opposite / Hypotenuse
• cos(angle) = Adjacent / Hypotenuse
• tan 30° = 1/√3 ≈ 0.577
• tan 45° = 1
• tan 60° = √3 ≈ 1.732
• tan 90° = undefined

In a right-angled triangle:

Key standard values to memorise:

4.Exam Patterns

What examiners ask — read before attempting PYQs

NDA regularly asks: 1. Find the height of a tower given distance and angle of elevation. 2. Two observers or two angles from the same base line (combined problems). 3. Problems involving a moving observer — angle changes as person walks closer. 4.

Shadow-based problems using angle of elevation of the sun. SHORTCUT / TRICK Trick 1 — The 45° Shortcut: When the angle of elevation is 45°, tan 45° = 1. So Height = Distance from base. No calculation needed! Trick 2 — Two-angle formula (very common in NDA): If a person stands at point A and walks distance 'd' closer to a tower, and the angle changes from α to β, then: Height of tower = d × tan α × tan β / (tan β − tan α) Memoise this formula. It saves 2-3 minutes in exam.

5.Worked Example

Solve this step-by-step before moving on
1
Step 1

Draw a right triangle. Let height = h, base = 40 m, angle = 30°.

2
Step 2

Use tan(angle) = height / base tan 30° = h / 40

3
Step 3

Substitute value of tan 30° = 1/√3 1/√3 = h / 40

4
Step 4

Solve for h h = 40 / √3 h = 40√3 / 3 ≈ 23.09 m Answer: Height of tower = 40/√3 = 40√3/3 metres

6.Exam Traps

Common mistakes students make — avoid these

Students confuse angle of elevation with angle of depression and set up the triangle incorrectly. Always draw the figure first. Mark the horizontal line clearly.

The angle is always measured FROM the horizontal, NOT from the vertical. Also, never forget to rationalise the denominator in your final answer.

Key Points to Remember

  • Angle of elevation is measured upward from the horizontal line to the line of sight.
  • Angle of depression is measured downward from the horizontal line to the line of sight.
  • Angle of elevation from point A to B always equals angle of depression from B to A.
  • Formula: tan(angle) = Height / Base Distance — this is the most used formula in this topic.
  • When angle of elevation = 45°, then Height = Base Distance (since tan 45° = 1).
  • Two-angle formula: Height = d × tan α × tan β / (tan β − tan α), where d is the distance walked.
  • tan 30° = 1/√3, tan 45° = 1, tan 60° = √3 — these three values cover 90% of exam questions.
  • Always draw a diagram first — it prevents wrong triangle setup and saves marks.

Exam-Specific Tips

  • tan 45° = 1, which means at 45° angle of elevation, height of tower equals the horizontal distance from the tower.
  • tan 60° = √3 ≈ 1.732 — used when the angle of elevation is 60°, giving height = √3 × base distance.
  • The angle of depression from the top of a tower equals the angle of elevation from the ground to the top — both are alternate interior angles.
  • Two-angle height formula: Height = (d × tan α × tan β) / (tan β − tan α), where d = distance between two observation points.
  • tan 30° = 1/√3 = √3/3 — rationalised form is √3/3, which must be used in final simplified answers.
  • In shadow problems, the angle of elevation of the sun equals the angle between the shadow (ground) and the line from tip of shadow to top of object.
  • If a tower and its shadow are equal in length, the angle of elevation of the sun is exactly 45°.
  • sin 30° = 1/2, cos 60° = 1/2, sin 45° = cos 45° = 1/√2 — these appear in inclined plane and rope-type distance problems.
Practice MCQs

Heights & Distances — Practice Questions

28graded MCQs · easy to hard · full solution & trap analysis · showing 20 of 28

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Practice 1easy

From a point on level ground, the angle of elevation to the top of a building is 45°. After walking 50 m towards the building, the angle of elevation becomes 60°. Find the height of the building.

Practice 2easy

A man standing at a point P observes the angle of elevation to the top of a vertical tower as 30°. He walks 100 m towards the base of the tower along level ground, and the angle of elevation becomes 60°. Find the height of the tower.

Practice 3easy

A person standing on the ground observes the angle of elevation to the top of a tree to be 45°. After walking 10 m towards the tree, the angle of elevation becomes 60°. Find the height of the tree.

Practice 4easy

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

Practice 5easy

Two vertical poles of heights 10 m and 15 m stand on level ground 12 m apart. Find the angle of elevation from the top of the shorter pole to the top of the taller pole.

Practice 6easy

A ladder leans against a vertical wall. The angle between the ladder and the ground is 60°. If the ladder is 10 m long, find the height at which the top of the ladder touches the wall.

Practice 7easy

A man standing at a point P on level ground observes the angle of elevation to the top of a vertical pole to be 30°. He walks 20 m directly towards the base of the pole and observes the angle of elevation to be 60°. Find the height of the pole.

Practice 8easy

From the top of a cliff 100 m 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 9easy

Two buildings of heights 20 m and 30 m stand on level ground 50 m apart. Find the angle of elevation from the top of the shorter building to the top of the taller building.

Practice 10easy

A ladder 13 m long leans against a vertical wall. If the angle between the ladder and the ground is 60°, how high up the wall does the ladder reach?

Practice 11medium

From the top of a building, the angles of depression to the top and bottom of a vertical pole standing on the same level ground are 30° and 60° respectively. If the building is 20 m tall, find the height of the pole.

Practice 12medium

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

Practice 13medium

A person standing at point A on level ground observes the angle of elevation to the top of a vertical tower at point B as 30°. After walking 40 m towards the base of the tower (to point C), the angle of elevation becomes 60°. Find the height of the tower.

Practice 14medium

From the top of a cliff of height 100 m, the angle of depression to a boat at sea is 30°. A second boat is directly behind the first, and the angle of depression to the second boat is 60°. Find the distance between the two boats (assume they are on the same horizontal line).

Practice 15medium

A vertical pole of height 12 m stands on level ground. An observer at point P on the ground sees the top of the pole at an angle of elevation α, where tan(α) = 3/4. The observer then moves 5 m further away from the pole (to point Q), and the angle of elevation becomes β. If tan(β) = 1/2, verify which of the following is the distance PQ.

Practice 16medium

Two buildings of heights h₁ = 20 m and h₂ = 30 m stand on level ground, separated by a horizontal distance of 15 m. An observer at the top of the first building looks at the top of the second building. Find the angle of elevation from the top of the first building to the top of the second building.

Practice 17medium

From a point A on the ground, the angle of elevation to the top of a tower is 45°. From a point B on the ground, 20 m away from A (in the direction away from the tower), the angle of elevation is 30°. Find the height of the tower.

Practice 18medium

A man standing at a point P observes the angle of elevation to the top of a vertical tower AB to be 30°. He then walks 40 m towards the base of the tower along level ground and observes the angle of elevation to be 60°. Find the height of the tower.

Practice 19medium

From the top of a cliff 100 m high, the angles of depression to two boats on the water are 30° and 60° respectively. If the boats are on the same side of the cliff and in line with the foot of the cliff, find the distance between the boats.

Practice 20medium

A vertical pole of height 12 m stands on level ground. An observer at a point on the ground 5 m away from the base of the pole observes the angle of elevation to the top of the pole as α. Later, the observer moves 7 m further away (along the same line) and observes the angle of elevation as β. If tan(α − β) = 1/11, find tan(α).

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60-Second Revision — Heights & Distances

  • Remember: tan(angle) = Height / Distance — this single formula solves most Heights & Distances problems.
  • Shortcut: Angle = 45° means Height = Distance from base. No calculation needed.
  • Formula: Two-angle problems → Height = d × tan α × tan β / (tan β − tan α). Memorise this.
  • Trap: Do NOT confuse angle of elevation (look up) with angle of depression (look down). Draw the figure every time.
  • Remember: Angle of elevation from ground = Angle of depression from top. They are always equal.
  • Values: tan 30° = 1/√3, tan 45° = 1, tan 60° = √3. These cover almost every NDA question.
  • Trap: Always rationalise the denominator in your final answer (e.g., 40/√3 must be written as 40√3/3).
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