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 = 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.