Some Applications of Trigonometry

Why This Chapter Matters

This is a standalone chapter that applies Chapter 8's concepts to real-world height and distance problems. Every board exam has 3-5 marks from this chapter. The problems follow a predictable pattern — master the key formulas and diagram-drawing, and this chapter is easy marks.

Prerequisites

Core Concepts

1. Key Definitions

Angle of Elevation: The angle formed at the observer's eye when looking UP at an object above the horizontal.

Angle of Depression: The angle formed at the observer's eye when looking DOWN at an object below the horizontal.

                  Object (TOP)
                    /
                   / ← Angle of elevation from A
                  /
Horizontal ______/________________
         A (Observer)

Horizontal ______________________
         B (Observer)
                  \
                   \ ← Angle of depression from B
                    \
                  Object (BOTTOM)

Key Fact: Angle of elevation from A to B = Angle of depression from B to A (alternate interior angles when horizontal lines are parallel)

2. Method for Solving

Step 1: Draw a clear, labelled diagram

Step 2: Identify the right triangle(s)

Step 3: Write tan (or other ratio) for the given angle

Step 4: Solve for the unknown

Step 5: Always write the final answer with units

Solved Examples

Example 1 — Single Triangle

Q: A tower stands vertically. From a point 20m away, the angle of elevation of the top is 60°. Find the height.

Let height = h

tan 60° = h/20

√3 = h/20

h = 20√3 metres

Example 2 — Two Triangles

Q: 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 the foot is 45°. Find the height of the tower.

Let: building height = AB = 7m, tower height = CD, horizontal distance = BC = d

Angle of depression of foot C from top A = 45°:

tan 45° = AB/BC → 1 = 7/d → d = 7m

Let ED = height of tower above building level (where E is at same level as A)

Angle of elevation of top D = 60°:

tan 60° = ED/AE = ED/d → √3 = ED/7 → ED = 7√3

Total height of tower = CD = ED + DC = ED + AB = 7√3 + 7 = 7(√3 + 1) metres

Example 3 — Moving Observer

Q: A man observes a tower from two points A and B, 100m apart on the same horizontal. Angles of elevation are 30° and 45°. Find height of tower.

Let tower height = h, distance from B to base = x

From B: tan 45° = h/x → x = h

From A: tan 30° = h/(100 + x) = h/(100 + h)

1/√3 = h/(100 + h)

100 + h = h√3

100 = h(√3 − 1)

h = 100/(√3 − 1) = 100(√3 + 1)/2 = 50(√3 + 1) metres

PYQs

2023

Q: From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed at the top of a 20m high building are 45° and 60°. Find the height of the tower.

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

From ground to top of tower: tan 60° = (20+h)/20 → √3 = (20+h)/20

20√3 = 20 + h → h = 20(√3 − 1) metres

2022

Q: The angle of elevation of the top of a hill from the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50m high, find the height of the hill.

Let hill height = h, horizontal distance = d

From tower foot: tan 60° = h/d → h = d√3 ... (1)

From hill foot: tan 30° = 50/d → d = 50√3 ... (2)

From (1) and (2): h = 50√3 × √3 = 150m

2021

Q: A ladder leaning against a wall makes 60° with ground. Foot is 2.5m from wall. Find length of ladder.

cos 60° = 2.5/L → 1/2 = 2.5/L → L = 5m

2020

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

Let point be at distance x from one pole and (80−x) from other. Height = h.

tan 60° = h/x → h = x√3

tan 30° = h/(80−x) → h = (80−x)/√3

x√3 = (80−x)/√3 → 3x = 80−x → 4x = 80 → x = 20

h = 20√3 metres

MCQ Practice

Q1. If the shadow of a pole is √3 times its height, the angle of elevation of the sun is:

(A) 60° (B) 45° (C) 30° ✓ (D) 90°

[tan θ = h/(√3h) = 1/√3 → θ = 30°]

Q2. From a cliff 200m high, angle of depression of a boat is 30°. Distance of boat from cliff base:

(A) 200/√3 (B) 200√3 ✓ (C) 400 (D) 100

[tan 30° = 200/d → d = 200√3]

Q3 (Hard). The angle of elevation of the top of a tower from point A is 30°. Walking 40m towards the tower, the angle is 60°. Find height.

Let h = height, x = distance from second point.

tan 60° = h/x → x = h/√3

tan 30° = h/(40+x) → 40+x = h√3 → 40 = h√3 − h/√3 = 2h/√3

h = 20√3 metres

Revision Notes

Angle of Elevation → looking UP → at observer's level below object
Angle of Depression → looking DOWN → at observer's level above object

Angle of elevation from A = Angle of depression from B (when horizontal)

tan(angle) = opposite/adjacent
→ Always: tan(elevation) = height/horizontal distance

Common values:
  tan 30° = 1/√3     tan 45° = 1     tan 60° = √3

For single triangle: one equation, one unknown
For two triangles: two equations, solve simultaneously

Common Mistakes:

❌ Not drawing a diagram — ALWAYS draw and label first

❌ Confusing elevation and depression

❌ Using sin/cos when tan gives the most direct relationship for height/distance

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