Question

From the top of a vertical cliff 40 m high, the angle of depression of an object that is level with the base of the cliff is 34º. How far is the object from the base of the cliff?

Answer 1

Answer:

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Answer 2

Recall:

$\sin \theta = \dfrac{\text{opposite}}{\text{hypotenuse}}$

$\cos \theta = \dfrac{\text{adjacent}}{\text{hypotenuse}}$

$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

* See attached for a visual representation of the question.

Find the distance between the cliff and the object:

Let D be the distance between the cliff and the object

$\tan \theta = \dfrac{\text{opposite}}{\text{adjacent}}$

$\tan (34) ^\circ = \dfrac{\text{40}}{\text{D}}$

$\text {D} = \dfrac{\text{40}}{\tan (34) ^\circ}$

$\text {D} = 59.3 \text { m}$

Answer: The object is 59.3m away from the cliff.