Question

Two poles of equal heights are standing opposite to each other on either side of the road which is 80 m wide. From a point P between them on the road, the angle of elevation of the top of one pole is 60° and the angle of depression from the top of another pole at P is 30°. Find the height of each pole and distances of the point P from the poles.

Answer 1

Let AB and CD be the two poles of equal height standing on the two sides of the road of width 80 m.

AC = 80 m

Let AP = x then PC = 80 – x

P is a point on the road from which the angle of elevation of the top of tower AD is 60°, also, the angle of depression of the point P from the point B is 30°.

∠BPA = 60° and ∠DPC = 30°

Draw a parallel line from DS to CA.

∠SDP = ∠DPC = 30°.

To Find : The height of the poles and the distance of the point P from both the poles.

From right △PAB:

tan 60° = AB/AP

√3 = h/x

or h = x√3 …(1)

From right △PCD:

tan 30° = AB/AP

1/√3 = h/(80-x)

or h = (80-x)/√3 …(2)

Form (1) and (2), we have

(80-x)/√3 = x√3

80 – x = 3x

x = 20

Form (1): h = (20)√3

Therefore,

Height of each pole = (20)√3 m

Distance of pole AB from point P = 20 m

Distance of pole CD from point P = 80 – 20 = 60 m