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A boy $1.7$ m tall, is $25$ m away from a tower and observes the angle of elevation of the top of the tower to be $60^{o}$ . What is the height of the tower?

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An Intiative by Sri Chaitanya

55

45

20

35

answer is B.

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Detailed Solution

Given that,
Height of the boy = 1.7m,
Distance of the boy from the tower = 25m,

θ = 60^{o}

Considering

A A^{'}

= the boy and the tower =

B^{'} C

.
Then,

A A^{'} = 1.7

= B B^{'}

A^{'} B^{'} = 25 m = AB

We have the formula,

tanθ = \frac{Height}{distance}

We have to find the height,

\Rightarrow tanθ = \frac{BC}{AB}

\Rightarrow 25 \tan 60^{o} = BC

\Rightarrow BC = 25 \sqrt{3}

Height of tower, h will be

\Rightarrow h = BC + B B^{'}

\Rightarrow

h = 25 \sqrt{3} + 1.7

\Rightarrow

h = 45

m (

\sqrt{3} = 1.732)

Thus, the height is 45m.
Therefore the correct option is 2.

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