sir this is the question from the book -ML aggarwal( avichal publication) class 10th , chapter20 , heights and distances

we have the inoformation about the question that A vertical tower is 20 m high.

A man standing at some distance from the tower knows that

the cosine of the angle of elevation of the top of the tower is 0.53.

How far is he standing from the foot of the tower?

an important question from the chapter

question no 11 , heights and distances , ICSE board, ML Aggarwal

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# A vertical tower is 20 m high. A man standing at some distance from the tower knows that the cosine of the angle of elevation of the top of the tower is 0.53. How far is he standing from the foot of the tower?

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Consider AB as the tower

Take a man C stands at a distance x m from the foot of the tower

cos θ = 0.53

We know that

Height of the tower AB = 20 m

cos θ = 0.53

So we get

θ = 58

^{0}Let us take

tan θ = AB/CB

Substituting the values

tan 58

^{0}= 20/xSo we get

1.6003 = 20/x

By cross multiplication

x = 20/1.6003

x = 12.49 = 12.5 m

Hence, the height of the tower is 12.5 m.