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.

The Angles Of Depression Of Two Objects From The Top Of A 10...

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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