Sec 3 · G1 (N(T))

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)

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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)

Practise this →