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An observer $1.5 m$ tall is $20.5 m$ away from a tower $22 m$ high. Determine the angle of elevation of the top of the tower from the eye of the observer.

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45 °

55 °

49 °

47 °

answer is A.

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Detailed Solution

It is given that an observer

1.5 m

tall is

20.5 m

away from a tower

22 m

high.
Drawing the required figure,
Question Image

In a right-angle triangle,

\tan x = \frac{Opposite side}{Adjacent side}

We have to find the angle of elevation of the top of the tower from the eye of the observer.
The height of the observer is 15m.
So,

AE = 22 - 1.5

AE = 20.5 m

In right angle

∆

AED,

\tan \tan x = \frac{AE}{ED}

\Rightarrow \tan \tan x = \frac{20.5}{20.5}

\Rightarrow \tan \tan x = 1

\Rightarrow x = 45 °

Thus, the angle of elevation is

x = 45 °

.
Hence, the correct option is (1).

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