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