8 min readConcept sheetPart II (2 marks); Part III (5 marks) when the observer's height is given

Heights and distances: angle of elevation

When you look up at an object, the angle between the horizontal line and your line of sight is the angle of elevation. Use tan θ = opposite/adjacent.

Line of sightThe line from the observer's eye to the object

Angle of elevationThe angle between the horizontal line and the line of sight, when the object is above the eye

Ratio used mosttan θ = opposite/adjacent = height/distance

Valuestan 30° = 1/√3, tan 45° = 1, tan 60° = √3; take √3 = 1.732

Key words

The observer is the person looking. The line of sight joins the observer's eye to the object. When the object is above the eye, the angle between the horizontal line and the line of sight is the angle of elevation. An instrument called a clinometer measures this angle. Towers, trees and walls are taken to stand vertically, so they make a right angle with the level ground.

Values you need

θ	sin θ	cos θ	tan θ
30°	1/2	√3/2	1/√3
45°	1/√2	1/√2	1
60°	√3/2	1/2	√3

How to draw the figure

1
Draw a horizontal line for the ground.
2
Draw a vertical line for the tower or tree. Mark its top A and its foot B.
3
Mark the observer's point C on the ground (or a horizontal line at eye level, if the observer's height is given).
4
Join C to A. This is the line of sight.
5
Mark the angle of elevation at C, between the horizontal line and CA. Mark the right angle at B.
6
Write the known lengths, and use h for the unknown height.

Example 1: height of a temple gopuram

Problem

From a point 40 m away from the foot of a temple gopuram, the angle of elevation of its top is 60°. Find the height of the gopuram. (√3 = 1.732)

Working

Formula: tan θ = opposite side/adjacent side
Let AB = h be the gopuram and C the point, with BC = 40 m and ∠ACB = 60°.
tan 60° = AB/BC = h/40
√3 = h/40
h = 40√3 = 40 × 1.732 = 69.28

Answer

The gopuram is 69.28 m high.

Example 2: when the observer's height counts

Problem

A boy whose eyes are 1.2 m above the ground stands 18 m from a coconut tree. He sees the top of the tree at an angle of elevation of 30°. Find the height of the tree. (√3 = 1.732)

Working

Formula: tan θ = opposite side/adjacent side
Draw a horizontal line from the boy's eye E to the tree, meeting it at D. ED = 18 m.
Let AD = x be the part of the tree above eye level.
tan 30° = x/18, so 1/√3 = x/18
x = 18/√3 = 18√3/3 = 6√3 = 6 × 1.732 = 10.392 m
Height of the tree = x + 1.2 = 10.392 + 1.2 = 11.592 m

Answer

The tree is about 11.59 m high (rounded to two decimal places).

In real life

Surveyors use a clinometer to find the height of a temple tower, a hill or a cell tower without climbing it. They measure one angle and one distance on the ground.

Exam angle

2 marks: “The angle of elevation of the top of a tree from a point 20 m from its foot is 45°. Find the height of the tree.” tan 45° = h/20, so 1 = h/20 and **h = 20 m**.

2 marks: “Find the angle of elevation of the sun when a 6 m pole casts a shadow 2√3 m long.” tan θ = 6/(2√3) = 3/√3 = √3, so **θ = 60°**.

5 marks: “A man 1.6 m tall stands 25 m from a cell tower. The angle of elevation of the top of the tower from his eyes is 60°. Find the height of the tower. (√3 = 1.732)” Above eye level: tan 60° = x/25, so x = 25√3 = 43.30 m. Height of the tower = 43.30 + 1.6 = **44.90 m**.

Remember it

TOA: Tan = Opposite/Adjacent. The height is opposite the angle; the ground distance is adjacent.