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 = 580
Let us take
tan θ = AB/CB
Substituting the values
tan 580 = 20/x
So 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.