In the adjoining figure, the angle of elevation of the top P of a vertical tower from a point X is 60 degree; at a point Y, 40 m vertically above X, the angle of elevation is 450. Find (i) the height of the tower PQ (ii) the distance XQ (Give your answer to the nearest metre)
In the adjoining figure, the angle of elevation of the top P of a vertical tower from a point X is 60 degree; at a point Y, 40 m vertically above X, the angle of elevation is 450. Find (i) the height of the tower PQ (ii) the distance XQ (Give your answer to the nearest metre)

Solution:

Consider PQ as the tower = h

XQ = YR = y

XY = 40 m

PR = h – 40

In right triangle PXQ

tan θ = PQ/XQ

Substituting the values

tan 600 = h/y

So we get

√3 = h/y

y = h/√3 ….. (1)

In right triangle PYR

tan θ = PR/YR

Substituting the values

tan 450 = (h – 40)/ y

So we get

1 = (h – 40)/ y

y = h – 40 …… (2)

Using both the equations

h – 40 = h/√3

By further calculation

√3h – 40√3 = h

√3h – h = 40√3

So we get

(1.732 – 1)h = 40 (1.732)

732h = 69.280

By division

h = 69.280/0.732 = 69280/732 = 94.64

Here

Height of the tower = 94.64 m = 95 m

Distance XQ = h – y = 95 – 40 = 55 m