Question

From a building 60m high,the angle of depression of top & bottom of the lamp post are 60dgree nd 30degree respectively.Find the distance between the lamp post

Answer 1

Given:- the angle of depression from a building with height 60m to the lamp post are 60° and 30° respectively as shown in the figure attached.

To find:- The distance between the lamp post and building*.

Formula used:-

$\tt tan60\degree= \dfrac{perpendicular}{base}$

$\tt tan60\degree= \sqrt{3}$

Solution :-

Let distance between B and C be x meter.

✏ In the △ABC,

$\sf \implies \: tan60 \degree= \dfrac{perpendicular}{base}$

$\sf \implies\tan60 \degree = \dfrac{60}{x}$

$\sf \implies \sqrt{3} = \dfrac{60}{x}$

$\sf \implies \: x = \dfrac{60}{ \sqrt{3} } m$

Therefore, the distance between lamp post and building is 60/√3m.

Answer 2

Step-by-step explanation:

Refer the attachment for figure.

From a building 60m high,the angle of depression of top & bottom of the lamp post are 60° and 30° respectively.

As per given condition,

In ∆CDE

tan30° = (60 - h)/x

1/√3 = (60 - h)/x

x = √3(60 - h) .............(1)

In ∆ABC

tan60° = 60/x

√3 = 60/x

x = 60/√3 .............(2)

On comparing (1) & (2) we get,

√3(60 - h) = 60/√3

3(60 - h) = 60

180 - 3h = 60

3h = 120

h = 40

Hence, the distance between the lamp post is 60/√3 m.