The angle of elevation of the top of a tower 30 m high from the foot

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Q) The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60⁰ and the angle of elevation of the top of the second tower from the foot of the first tower is 30⁰. Find the distance between the two towers and also the height of the other tower.

Ans:

Let AB and PQ be the towers, heights be 30 m & H m; distance between 2 towers be D m, and angle of elevations be as shown in the image above.

In Δ ABQ, $\tan 60 = \frac{AB}{BQ}$

$\sqrt 3 = \frac{30}{D}$

$\therefore D = \frac{30}{\sqrt 3}$

or D = 10√3 m

Now in Δ PQB, $\tan 30 = \frac{PQ}{QB}$

$\frac{1}{\sqrt 3} = \frac{H}{10\sqrt3}$

$\therefore$ H = $\frac{10\sqrt3}{\sqrt 3}$

or H = 10 m

Therefore, the distance between the two towers is 10√3 m and the height of the other tower is 10 m.

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