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Question

An observer 1.5metres tall is 20.5metres away from a tower 22metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.


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Solution

Given, the length of the observer is 1.5metres and the height of the tower is 22metres.

The observer is 20.5metres away from a tower.

In the figure, let CD represents the height of the observer, AB represents the height of the tower and AC represents the distance of the observer from the tower.

Then AB=22m,CD=1.5m,AC=20.5m

From the figure, BE=AB-AE=22-1.5=20.5m and DE=20.5m

Let θ be the angle of elevation of the top of the tower from the eye of the observer.

In BED,

tanθ=BEDE=20.520.5=1..............(1) (tanθ=perpendicularbase)

Since we know that by the trigonometry values,

tan45°=1..............(2)

From equations (1) and (2), we get;

tanθ=tan45°θ=45°

Therefore the angle of elevation is 45°.


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