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

The angle of elevation of the top of an unfinished tower

at a point distant 120 m from its base is 45.

How much higher must the tower be raised

so that its angle of elevation at the same point may be 60?

question no 36 , heights and distances , ICSE board, ML Aggarwal

Consider AB as the unfinished tower where AB = 120 m

Angle of elevation = 45

^{0}Take x be higher raised so that the angle of elevation becomes 60

^{0}BC = y

In right triangle ABC

tan θ = AB/CB

Substituting the values

tan 45

^{0}= AB/CB = 120/ySo we get

1 = 120/y

y = 120 m

In right triangle DBC

tan 60

^{0}= DB/CBSubstituting the values

√3 = (120 + x)/ 120

120√3 = 120 + x

x = 120√3 – 120

x = 120 (√3 – 1)

So we get

x = 120 (1.732 – 1)

x = 120 (0.732)

x = 87.84 m

Hence, the tower should be raised at 87.84 m.