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An observer $1.5$ m tall is $28.5$ m away from a tower and the angle of elevation from the eye of the observer is $45^{o}$ . The height of the tower is?

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27

30

28.5

None of these

answer is B.

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Detailed Solution

It is given that an observer

1.5

m tall is

28.5

m away from a tower and the angle of elevation from the eye of the observer is

45^{o}

.
Let

AB

be the observer and

CD

be the tower.
Question Image

AC = BE = 28.5 m

AB

CE =

1.5

m
Let

DE

h

m.
From right angled

△ BED

, tan

45^{o} = \frac{DE}{BE}

⇒

1 = \frac{h}{28.5}

⇒

h = 28.5

⇒

h + 1.5 = 28.5 + 1.5

⇒

h + 1.5 = 30

∴

Height of tower

CD = 30

m.
Hence, the correct option is 2.

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