The angles of depression of two objects from the top of a 100 m hill lying to it

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Question

The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be $45^{\circ} a n d 30^{\circ}$ . Find the distance between the two objects. (Take $\sqrt{3} = 1.732$ )

Options:

A . 73.2 metres

B . 107.5 meters

C . 150 metres

D . 200 metres

Answer: Option A
:
A

Let C and D be the objects and CD be the distance between the objects.

In $Δ A B C$ , $t a n 45^{\circ}$ = $\frac{A B}{A C}$ = 1

$A B = A C = 100 m$

In $Δ A B D$ , $t a n 30^{\circ}$ = $\frac{A B}{A D}$
$A D \times \frac{1}{\sqrt{3}} = 100$
$A D = 100 \times \sqrt{3} = 173.2 m$
$C D = A D - A C = 173.2 - 100 = 73.2 m e t r e s$

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