Question

The horizontal distance between two towers is 60 meter. The angel of depression of the top of the first tower when seen from the top of the second tower is 30°. If the height of the second tower is 90 meter. Find the height of the first tower

Answer 1

SOLUTION:-

Assume AB & CD be the two towers.

The height of the second tower is AB= 90m.

The horizontal distance between the two towers is BD= 60m.

&

The angles of depression of the first tower as seen from the top of the second tower is angle ACE=30°

⚫BD= AE= 60m

⚫AB= DE= 90m

Therefore,

In ∆ACE,

$tan30 \degree = \frac{CE}{AE} \\ \\ = > \frac{1}{ \sqrt{3} } = \frac{CE}{60} \\ \\ = > \sqrt{3} CE = 60 \\ \\ = > CE = \frac{60}{ \sqrt{3} } \\ [Rationalising] \\ = > \frac{60 \times \sqrt{3} }{ \sqrt{3} \times \sqrt{3} } \\ \\ = > \frac{60 \sqrt{3} }{3} \\ \\ = >CE = 20 \sqrt{3} m$

Therefore,

•Height of the First Tower;

CD= CE + DE

CD= (20√3 + 90)m

CD= (20×1.732+90)m

CD= (34.64 + 90)m

CD= 124.64m

Thus,

The height of the first tower is 124.64m.

Hope it helps ☺️