Question

Two poles are standing opposite to each other on the either side of the road which is 90 feet wide.The angle of elevation from bottom of the first pole to the top of the second pole is 45°.The angle of elevation from the bottom of the second pole to the top of the first pole is 30°Find the heights of the poles.(use√3=1.732)


​

Answer 1

Let the two poles be AB and CD

So, length of pole = AB =CD

length of road = 90 m

So, BC = 90 m

let point P be a point on the road between the poles

poles are perpendicular to ground

∠ABP = 90° , ∠DCP =90°

IN a right Δ DPC

tan P = CD/CP

tan 30°= CD/CP

1/ √3 = CD/CP

CP / √3 = CD ...(1)

in right Δ APB

tan P = AB/PB

tan 45° = AB /PB

1× PB = AB

PB =AB

CD = BP...(2)

From (1) & (2)

CP/ √3=BP

CP = BP / √3

Now BC = BP + CP

90 = BP + BP/√3

90 = BP + 1.732 BP

90 = 2.732 BP

BP = 32.9 m

Now CP = BC- BP

CP = 90 - 32.9

CP = 57.1 m

From CD = BP

so CD = 32.9m