The angle of elevation of the top of an unfinished tower at a point distant 120 m from its base is 45⁰. How much higher must the tower be raised so that its angle of elevation at the same point may be 60⁰?

Solution:

Consider AB as the unfinished tower where AB = 120 m

Angle of elevation = 45⁰

Take x be higher raised so that the angle of elevation becomes 60⁰

BC = y

In right triangle ABC

tan θ = AB/CB

Substituting the values

tan 45⁰ = AB/CB = 120/y

So we get

1 = 120/y

y = 120 m

In right triangle DBC

tan 60⁰ = DB/CB

Substituting 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.