Question

A pole 5m high is fixed on the top of the tower. The angle of elevation of the top of the pole observed from a point 'A' on the ground is 60° and the angle of depression of the pont 'A' from the top of the tower is 45°. Find the height of the tower

Answer 1

ANSWER:-

Given:

A pole 5m high is fixed on the top of the tower. The angle of elevation of the top of the pole observed from a point A on the ground is 60° and the angle of depression of the point A from the top of the tower is 45°.

To find:

Find the height of the tower.

Solution:

angle BAD=60°

angle BAC=45°

In right ∆ABC,

$tan45 \degree = \frac{BC}{AB} \\ \\ = > 1 = \frac{h}{AB} \\ \\ = > AB = h...................(1)$

&

In right ∆ABD,

$tan60 \degree = \frac{BD}{AB} \\ \\ = > \sqrt{3} = \frac{BC + CD}{AB} \\ \\ = > \sqrt{3} = \frac{h + 5}{AB} \\ \\ = > \sqrt{3} AB = h + 5 \\ \\ = > AB = \frac{h + 5}{ \sqrt{3} }...............(2)$

Comparing equation (1) & (2), we get;

$= > h = \frac{h + 5}{ \sqrt{3} } \\ \\ = > \sqrt{3}h = h + 5 \\ \\ = > \sqrt{3} h - h = 5 \\ \\ = > h( \sqrt{3} - 1) = 5 \\ \\ = > h = \frac{5}{ \sqrt{3} - 1} \times \frac{ \sqrt{3} + 1}{ \sqrt{3} + 1 } \\ \\ = > h = \frac{5( \sqrt{3} + 1)}{({ { \sqrt{3}) }^{2} - ( {1)}^{2} } } \\ \\ = > h = \frac{5( \sqrt{3} + 1) }{3 - 1} \\ \\ = > h = \frac{5( \sqrt{3} + 1)}{2} \\ \\ = > h = \frac{5(1.732 + 1)}{2 } \\ \\ = > h = \frac{5 \times 2.732}{2} \\ \\ = > h = \frac{13.66}{2} \\ \\ = > h = 6.83m$

Hence,

Height of the tower be 6.83m

Hope it helps ☺️