Question

A man is standing on the top of a building 60 m high. He observes that the angle of depression of the top and bottom of tower has measure 30 and 60 respectively. Find the height of the tower​

Answer 1

Answer:

Hope it's help you

Answer 2

ANSWER:-

Given:

A man is standing on the top of a building 60m high. He observes that angle of depression of the top and bottom of tower has measure 30° & 60° respectively.

To find:

Find the height of the tower.

Solution:

⏺️Let AC be the tower and BE be the building.

⏺️Let height of the tower be h m.

It is given that the angles of depression of the top C and bottom A of the tower.

According to the question:

In right ∆CDE,we have;

$= > tan30 \degree = \frac{DE}{CD} \\ \\ = > \frac{1}{ \sqrt{ 3} } = \frac{60 - h}{CD} \: \: \: \: \: \: \: (DE = BE - BD = BE - AC)\\ \\ = > CD = \sqrt{3} (60 - h)...........(1)$

In ∆ABE, we have;

$= > tan60 \degree = \frac{BE}{AB} \\ \\ = > \sqrt{3} = \frac{60}{CD} \: \: \: (AB = CD) \\ \\ = > \sqrt{3} CD = 60 \\ \\ = > CD = \frac{60}{ \sqrt{ 3} } .............(2)$

Therefore,

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

=) √3(60-h) = 60/√3

=) 3(60-h) = 60

=) 180 -3h= 60

=) -3h= 60 -180

=) -3h= -120

=) h= -120/-3 [minus cancel]

=) h= 40m

Hence,

The height of the tower is 40m.

Hope it helps ☺️