Two poles of equa

Let AD and BC be the two poles of equal height standing on the two sides of the road of width 80 m. Join DC. Then DC = 80 m. 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°. Draw a line BX from B parallel to the ground. Then ∠XBP = ∠BPC = 30°. We are to find the height of the poles and the distance of the point P from both the poles. Also, ∠APD = 60°.

Let DP = x. Then PC = 80-x.

In ∆APD,

or,

Now, from ∆APD,

or,

Now, BC = AD

So,

or,

or,

x = 20 m

So,PD = 20 m. Hence, PC = 80-20 = 60 m. Also, BC = AD = 20√3 m.