Sec 3 · G3 (Express)

Applications of trigonometry

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1. Angles of elevation and depression

Left: looking up from the ground, the angle of elevation is between the horizontal and the line of sight. Right: looking down from the top of a cliff, the angle of depression is between the horizontal and the line of sight; it equals the angle of elevation from the boat.

Ratio	Formula
$\sin\theta$	$\dfrac{\text{opposite}}{\text{hypotenuse}}$
$\cos\theta$	$\dfrac{\text{adjacent}}{\text{hypotenuse}}$
$\tan\theta$	$\dfrac{\text{opposite}}{\text{adjacent}}$

Find a height: 20 m away, elevation 62°

\displaystyle \tan 62^\circ = \dfrac{h}{20}

e.g.

\displaystyle \tan 62^\circ = \dfrac{h}{20}

\displaystyle h = 20\tan 62^\circ

\displaystyle h = 37.6\text{ (3 s.f.)}

Find a distance: cliff 24 m, depression 18°

\displaystyle \tan 18^\circ = \dfrac{24}{x}

e.g.

\displaystyle \tan 18^\circ = \dfrac{24}{x}

\displaystyle x = \dfrac{24}{\tan 18^\circ}

\displaystyle x = 73.9\text{ (3 s.f.)}

Find the angle: 20 m tall, 45 m away

\displaystyle \tan\theta = \dfrac{20}{45}

e.g.

\displaystyle \tan\theta = \dfrac{20}{45}

\displaystyle \theta = \tan^{-1}\left(\dfrac{20}{45}\right)

\displaystyle \theta = 24.0^\circ\text{ (1 d.p.)}

📖 Worked example

From the top of a cliff $24$ m high, the angle of depression of a boat is $18^\circ$ . Find the distance of the boat from the foot of the cliff, correct to 3 significant figures.

Step 1 of 3

Mark the angle at B

Angle

TBF = 18^\circ

(alt. ∠s)

📖 Worked example

$A$ and $B$ are points on level ground in a straight line with the foot $F$ of a flagpole, with $AB = 15$ m. The angles of elevation of the top $T$ from $A$ and $B$ are $32^\circ$ and $45^\circ$ respectively. Find the height of the flagpole, correct to 3 significant figures.

Step 1 of 4

Let BF = x: which ratio links h and the ground distance in each triangle?

In triangle

BFT

h = x \tan 45^\circ

In triangle

AFT

h = (x + 15) \tan 32^\circ

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2. Bearings

Left: a North arrow at A; the bearing of B from A is the clockwise angle from North to AB, 054°. Right: the bearing of B from A is 136°; at B, the clockwise angle from North to BA is 316°.

Rule	Example
from North, clockwise	north-east is $045^\circ$
three figures	$54^\circ$ is written $054^\circ$
back bearing: $\pm 180^\circ$	$136^\circ + 180^\circ = 316^\circ$
north–south distance	$d\cos\theta$ , $\theta$ = acute angle to the N–S line
east–west distance	$d\sin\theta$ , $\theta$ = acute angle to the N–S line

9 km on a bearing of 066°: how far north?

\displaystyle \cos 66^\circ = \dfrac{y}{9}

e.g.

\displaystyle \cos 66^\circ = \dfrac{y}{9}

\displaystyle y = 9\cos 66^\circ

\displaystyle y = 3.66\text{ (3 s.f.)}

4 km west and 8 km north: the bearing

\displaystyle \tan\alpha = \dfrac{4}{8}

e.g.

\displaystyle \tan\alpha = \dfrac{4}{8}

\displaystyle \alpha = 26.57^\circ\text{ (2 d.p.)}

north-west of $A$ : bearing $= 360^\circ - 26.57^\circ$ , so $333.4^\circ$

📖 Worked example

The bearing of $B$ from $A$ is $136^\circ$ . Find the bearing of $A$ from $B$ .

Step 1 of 2

The north lines at A and B are parallel

Angle from south at

B

BA

= 136^\circ

(alt. ∠s)

📖 Worked example

$B$ is $4$ km west and $8$ km north of $A$ . Find the bearing of $B$ from $A$ , correct to 1 decimal place.

Step 1 of 2

The angle α between AB and the north–south line: which ratio?

\tan \alpha = \dfrac{4}{8}

\alpha = \tan^{-1}\left(\dfrac{4}{8}\right)

\alpha = 26.5650\ldots^\circ

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3. Three-dimensional problems

Cuboid ABCDEFGH, AB = 8 cm, BC = 7 cm, CG = 3 cm. The base diagonal AC and the space diagonal AG are drawn; the angle between AG and the base is angle GAC, with a right angle at C.

Base diagonal (right angle at B)

\displaystyle AC^{2} = AB^{2} + BC^{2}

e.g.

\displaystyle AC = \sqrt{8^{2} + 7^{2}}

\displaystyle AC = \sqrt{113}

Space diagonal (right angle at C)

\displaystyle AG^{2} = AC^{2} + CG^{2}

e.g.

\displaystyle AG = \sqrt{113 + 3^{2}}

\displaystyle AG = \sqrt{122}

\displaystyle AG = 11.0\text{ (3 s.f.)}

Angle between AG and the base: angle GAC

\displaystyle \tan\theta = \dfrac{CG}{AC}

e.g.

\displaystyle \tan\theta = \dfrac{3}{\sqrt{113}}

\displaystyle \theta = 15.8^\circ\text{ (1 d.p.)}

📖 Worked example

The diagram shows a cuboid with $AB = 8$ cm, $BC = 7$ cm and $CG = 3$ cm. Find the angle between $AG$ and the base $ABCD$ , correct to 1 decimal place.

Step 1 of 4

First the base diagonal AC (right angle at B)

AC^2 = 8^2 + 7^2

AC^2 = 113

AC = \sqrt{113}

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Sec 3 · G3 (Express)

Applications of trigonometry

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Start free Worksheet (PDF)

1. Angles of elevation and depression

Ratio	Formula
$\sin\theta$	$\dfrac{\text{opposite}}{\text{hypotenuse}}$
$\cos\theta$	$\dfrac{\text{adjacent}}{\text{hypotenuse}}$
$\tan\theta$	$\dfrac{\text{opposite}}{\text{adjacent}}$

Find a height: 20 m away, elevation 62°

\displaystyle \tan 62^\circ = \dfrac{h}{20}

e.g.

\displaystyle \tan 62^\circ = \dfrac{h}{20}

\displaystyle h = 20\tan 62^\circ

\displaystyle h = 37.6\text{ (3 s.f.)}

Find a distance: cliff 24 m, depression 18°

\displaystyle \tan 18^\circ = \dfrac{24}{x}

e.g.

\displaystyle \tan 18^\circ = \dfrac{24}{x}

\displaystyle x = \dfrac{24}{\tan 18^\circ}

\displaystyle x = 73.9\text{ (3 s.f.)}

Find the angle: 20 m tall, 45 m away

\displaystyle \tan\theta = \dfrac{20}{45}

e.g.

\displaystyle \tan\theta = \dfrac{20}{45}

\displaystyle \theta = \tan^{-1}\left(\dfrac{20}{45}\right)

\displaystyle \theta = 24.0^\circ\text{ (1 d.p.)}

📖 Worked example

From the top of a cliff $24$ m high, the angle of depression of a boat is $18^\circ$ . Find the distance of the boat from the foot of the cliff, correct to 3 significant figures.

Step 1 of 3

Mark the angle at B

Angle

TBF = 18^\circ

(alt. ∠s)

📖 Worked example

Step 1 of 4

Let BF = x: which ratio links h and the ground distance in each triangle?

In triangle

BFT

h = x \tan 45^\circ

In triangle

AFT

h = (x + 15) \tan 32^\circ

Practise this →

2. Bearings

Rule	Example
from North, clockwise	north-east is $045^\circ$
three figures	$54^\circ$ is written $054^\circ$
back bearing: $\pm 180^\circ$	$136^\circ + 180^\circ = 316^\circ$
north–south distance	$d\cos\theta$ , $\theta$ = acute angle to the N–S line
east–west distance	$d\sin\theta$ , $\theta$ = acute angle to the N–S line

9 km on a bearing of 066°: how far north?

\displaystyle \cos 66^\circ = \dfrac{y}{9}

e.g.

\displaystyle \cos 66^\circ = \dfrac{y}{9}

\displaystyle y = 9\cos 66^\circ

\displaystyle y = 3.66\text{ (3 s.f.)}

4 km west and 8 km north: the bearing

\displaystyle \tan\alpha = \dfrac{4}{8}

e.g.

\displaystyle \tan\alpha = \dfrac{4}{8}

\displaystyle \alpha = 26.57^\circ\text{ (2 d.p.)}

north-west of $A$ : bearing $= 360^\circ - 26.57^\circ$ , so $333.4^\circ$

📖 Worked example

The bearing of $B$ from $A$ is $136^\circ$ . Find the bearing of $A$ from $B$ .

Step 1 of 2

The north lines at A and B are parallel

Angle from south at

B

BA

= 136^\circ

(alt. ∠s)

📖 Worked example

$B$ is $4$ km west and $8$ km north of $A$ . Find the bearing of $B$ from $A$ , correct to 1 decimal place.

Step 1 of 2

The angle α between AB and the north–south line: which ratio?

\tan \alpha = \dfrac{4}{8}

\alpha = \tan^{-1}\left(\dfrac{4}{8}\right)

\alpha = 26.5650\ldots^\circ

Practise this →

3. Three-dimensional problems

Base diagonal (right angle at B)

\displaystyle AC^{2} = AB^{2} + BC^{2}

e.g.

\displaystyle AC = \sqrt{8^{2} + 7^{2}}

\displaystyle AC = \sqrt{113}

Space diagonal (right angle at C)

\displaystyle AG^{2} = AC^{2} + CG^{2}

e.g.

\displaystyle AG = \sqrt{113 + 3^{2}}

\displaystyle AG = \sqrt{122}

\displaystyle AG = 11.0\text{ (3 s.f.)}

Angle between AG and the base: angle GAC

\displaystyle \tan\theta = \dfrac{CG}{AC}

e.g.

\displaystyle \tan\theta = \dfrac{3}{\sqrt{113}}

\displaystyle \theta = 15.8^\circ\text{ (1 d.p.)}

📖 Worked example

The diagram shows a cuboid with $AB = 8$ cm, $BC = 7$ cm and $CG = 3$ cm. Find the angle between $AG$ and the base $ABCD$ , correct to 1 decimal place.

Step 1 of 4

First the base diagonal AC (right angle at B)

AC^2 = 8^2 + 7^2

AC^2 = 113

AC = \sqrt{113}

Practise this →

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