Question

A window in a building is at height of 10m from the ground. The angle of depression of a point A on the ground from the window is 30degree. The angle of elevation of the top of the building from the point A is 60degree. Find the height of the building.

Answer 1

Given- AC = 10m

Let the base AP = x and BC = y

=> In triangle APC

Tan30° = 10/x

1/√3 = 10/x

x = 10√3

=> In triangle APC

Tan60° = QS/x

√3 = 10+y/10√3

3 × 10√3 = 10+y

=> y = 30 - 10

y = 20m

Therefore, The height of the building:

=> QS = 20+ 10

QS = 30m

Hope you like my ans

Answer 2

SOLUTION

Let C be a Window , BC = 10m

Now, In ∆ABC,

=) tan 30° = CB/AB

=) 1/√3 = 10/ AB

=) AB= 10√3 m.......(1)

In ∆ABD,

tan 60° = BD/AB

=) √3= BD/ 10√3 {Using (1)}

=) BD= 10√3 ×√3

=) BD= 10×3 = 30m

hope it helps ✔️