Height and Distance

Height and Distance is the direct application of Trigonometry to real-world geometry. The chapter's reputation for difficulty is mostly diagram-shyness — drawing the right triangle is half the work; applying \(\tan\theta = \text{opposite}/\text{adjacent}\) is the other half.

This page is built from CDS Previous Year Questions across 2007–2023, plus NCERT Class 10 Applications of Trigonometry. Pair with Trigonometry.

What This Topic Covers

CDS scope: (1) angle of elevation — looking up; (2) angle of depression — looking down; (3) single-observer problems — find height or distance; (4) two-observer problems — same object from two different angles; (5) moving-observer problems — angle changes as observer moves; (6) two-object problems — heights or distances of two objects from one observer.

Exam Pattern & Weightage

Year / Paper	No.	Subtopics Tested
2007-II	2	Tower height from elevation
2008-I/II	3	Two-observer, depression
2009-II	2	Moving observer
2010-II	2	Tower with two distances
2013-I/II	3	Angle of depression, river
2014-I/II	3	Two-observer, two-tower
2015-I/II	3	Mixed
2016-I/II	3	Moving observer, two-tower
2017-I/II	3	Aircraft, river, mountain
2019-II	2	Tower height
2020-I/II	3	Mixed
2021-I/II	3	Two-observer, depression
2022-I / 2023-I	3	Mixed

Core Concepts

Angle of Elevation and Depression

Elevation: if you look up from a horizontal line to an object higher than your eye, the angle between your line of sight and the horizontal is the angle of elevation. Depression: if you look down from a horizontal line, the angle between your sight and the horizontal is the angle of depression. They are alternate angles — geometrically equal in pair setups.

Basic Setup

So \(h = d \tan\theta\) or \(d = h / \tan\theta = h \cot\theta\). Always draw a right triangle with the object at the top, observer at the bottom, and horizontal ground as the base.

Two-Observer Setup (Same Side)

An observer at horizontal distance \(d_1\) sees angle \(\alpha\); moves to \(d_2 < d_1\) (closer), sees angle \(\beta > \alpha\). The tower height \(h\):

Solve simultaneously: \(d_1 - d_2 = h(\cot\alpha - \cot\beta)\). If the movement is given, find \(h\).

Two-Observer Setup (Opposite Sides)

If the two observers are on opposite sides of the tower, the distance between them = sum of two horizontal distances: \(d_1 + d_2\). Otherwise difference: \(d_1 - d_2\).

Aircraft / Cloud / Bird Problems

Same logic with an extra elevation. If a bird is at height \(h_b\), and the observer's eye is at height \(h_e\), the effective height for the right triangle is \(h_b - h_e\).

Worked Examples

Example 1 — Tower Height (2007-II)

Q: From a point 50 m from the foot of a tower, the angle of elevation of the top is 60°. Find the tower's height.

• \(h = d \tan\theta = 50 \tan 60° = 50\sqrt{3} \approx 86.6\) m.

Example 2 — Two Distances, Same Tower (2010-II)

Q: A tower's elevation from a point is 60°. From a point 20 m further, the elevation is 30°. Find the tower's height.

• Let height = \(h\), first distance = \(d\). Then \(\tan 60° = h/d \implies h = d\sqrt{3}\).
• From farther point: \(\tan 30° = h/(d + 20) \implies h = (d + 20)/\sqrt{3}\).
• Equate: \(d\sqrt{3} = (d + 20)/\sqrt{3} \implies 3d = d + 20 \implies d = 10\). So \(h = 10\sqrt{3} \approx 17.3\) m.

Example 3 — Angle of Depression (2008-II)

Q: From the top of a 100 m cliff, the angle of depression of a boat is 30°. How far is the boat?

• Angle of depression of boat = angle of elevation of cliff top from boat = 30°.
• \(\tan 30° = 100/d \implies d = 100/\tan 30° = 100\sqrt{3} \approx 173.2\) m.

Example 4 — Aircraft (2017-I)

Q: An aeroplane is flying at 5000 m. From a point on the ground, the angle of elevation is 60°. How far is the plane horizontally from the observer?

• \(\tan 60° = 5000/d \implies d = 5000/\sqrt{3} \approx 2886.75\) m.

Example 5 — Two Towers (2014-I)

Q: Two towers stand on level ground. Their heights are 30 m and \(h\) m, separated by 40 m. The angle of elevation of the top of the taller from the base of the shorter is 60°. Find \(h\).

• From base of shorter (height 30), looking up to top of taller (height \(h\)): horizontal distance 40, vertical difference \(h - 30\) (if \(h > 30\)).
• \(\tan 60° = (h - 30)/40 \implies h - 30 = 40\sqrt{3} \implies h = 30 + 40\sqrt{3}\) m \(\approx 99.3\) m.

Example 6 — Moving Observer (2009-II)

Q: The angle of elevation of a tower from a point is 30°. The observer moves 50 m closer and the angle becomes 60°. Find the tower's height.

• This is the standard CDS double-angle problem. From farther: \(\tan 30° = h/(d + 50) = 1/\sqrt{3}\). From closer: \(\tan 60° = h/d = \sqrt{3}\).
• From first: \(h = (d + 50)/\sqrt{3}\). From second: \(h = d\sqrt{3}\). Equate: \(d\sqrt{3} = (d + 50)/\sqrt{3} \implies 3d = d + 50 \implies d = 25\).
• \(h = 25\sqrt{3} \approx 43.3\) m.

Example 7 — Opposite Sides (2013-II)

Q: A tower stands between two observers. From one, elevation is 30°; from the other, 60°. The two observers are 200 m apart. Find the height.

• Let \(d_1, d_2\) be the distances from the two observers, \(d_1 + d_2 = 200\).
• \(h = d_1 \tan 30° = d_1/\sqrt{3}\); \(h = d_2 \tan 60° = d_2 \sqrt{3}\).
• So \(d_1 = 3 d_2\). With \(d_1 + d_2 = 200\): \(d_2 = 50, d_1 = 150\).
• \(h = 50\sqrt{3} \approx 86.6\) m.

Exam Shortcuts (Pro-Tips)

Shortcut 1 — Draw the Triangle First

Before doing any arithmetic, sketch the situation. Vertical: tower height. Horizontal: ground distance. Angle: elevation or depression. Most CDS slip-ups come from skipping this step.

Shortcut 2 — 30° and 60° Together → 1:3 Distance Ratio

Whenever a problem has both 30° and 60° as angles in a two-observer setup, the distance ratio is 1:3 (or 3:1, depending on which is closer). Apply this directly.

Shortcut 3 — Depression = Elevation (Alternate)

The angle of depression from the top equals the angle of elevation from the bottom (alternate angles between parallel horizontals). Use whichever is simpler to compute.

Shortcut 4 — Standard Values Table

\(\tan 30° = 1/\sqrt{3}\), \(\tan 45° = 1\), \(\tan 60° = \sqrt{3}\). Memorise — these appear in every CDS H&D problem.

Shortcut 5 — Algebraic Setup with One Unknown

Pick one unknown (height or distance), write all conditions in terms of it. Most CDS H&D problems are one-equation one-unknown after the diagram is drawn.

Common Question Patterns

Pattern 1 — Single Observer, Find Height

Given horizontal distance and angle of elevation, find the tower's height. Apply \(h = d \tan\theta\).

Pattern 2 — Two Observers Same Side

Observer moves closer; angle changes. Set up two equations \(h = d_1 \tan\alpha = d_2 \tan\beta\); solve.

Pattern 3 — Two Observers Opposite Sides

Tower between two observers. Distance between them = \(d_1 + d_2\). Solve.

Pattern 4 — Angle of Depression

From cliff or tower top, look down. Depression = elevation from below.

Pattern 5 — Two Objects (Two Towers, Cloud and Plane)

Two objects in the diagram. Often the elevation of one and depression of another are given.

Preparation Strategy

Week 1. Master the diagram. Practice 20 problems where you only have to draw the triangle and write \(\tan\theta = h/d\). No solving needed.

Week 2. Two-observer problems and depression. Drill the 1:3 distance ratio for 30°-60° pairs. Practice moving-observer problems where the angle changes.

Mock testing. Use CDS mock tests. Common slip-ups: misreading "elevation" vs "depression", or forgetting to add the observer's eye height in some problems.

Cross-train with Trigonometry (the formulas) and Triangles and Properties (the geometry of right triangles).

Frequently Asked Questions

What is the difference between angle of elevation and angle of depression?

Elevation is the angle between the horizontal and the line of sight to an object above. Depression is the angle between the horizontal and the line of sight to an object below. They are alternate angles in standard CDS setups, so depression from above equals elevation from below.

What is the basic formula for tower height?

If horizontal distance is \(d\) and angle of elevation is \(\theta\): height = \(d \tan\theta\). Equivalently, \(d = h / \tan\theta = h \cot\theta\). Always set up the right triangle with vertical = height, horizontal = distance, hypotenuse = line of sight.

Why is the 1:3 distance ratio important?

For a 30°-60° two-observer problem (same side), \(\tan 30°/\tan 60° = (1/\sqrt{3})/\sqrt{3} = 1/3\). So the farther observer is 3 times the distance of the nearer. Recognise this and many problems collapse to one equation.

How do I set up a two-observer problem?

Draw two right triangles sharing the tower as their common vertical leg. The two observer positions give two horizontal distances. Write \(h = d_1 \tan\alpha = d_2 \tan\beta\). Use given conditions on \(d_1\) or \(d_2\) (often \(d_1 - d_2\) is the observer's movement).

What if the observers are on opposite sides of the tower?

Then the tower is between them, and the total separation = \(d_1 + d_2\) (not \(d_1 - d_2\) as in same-side case). Apply the standard equations and use the sum constraint.

What about aeroplanes or birds (objects in the air)?

Same logic. The "height" is the object's altitude minus the observer's eye height (if non-trivial). Some problems mention "from the ground", meaning eye height ≈ 0.

Which CDS Maths topics connect to Height and Distance?

Trigonometry — direct parent; H&D is its application. Triangles and Properties — the underlying geometry of right triangles. Lines and Angles — angle relations like alternate angles.

Related Topics

CDS MathsTrigonometry Functions and Identities CDS MathsTriangles and Properties CDS MathsLines and Angles CDS MathsCircles CDSCDS Mathematics — Full Syllabus