Question

From the top of a 7 m high building, the angle of elevation of the top of a cable is 60° and the angel of depression of its foot is 45°. Determine the height of the tower.

Answer 1

Given

Height of the building = 7 m = AB

Height of the cable tower = CD

Angle of elevation of the top of the cable tower from the top of the building = 60°

Angle of depression of the bottom of the building from the top of the building= 45°

Then, from the fig. we see that

ED = AB = 7 m

And,

CD = CE + ED

So, In ΔABD, we have

AB/ BD = tan 45^o

AB = BD = 7

BD = 7

In ΔACE,

AE = BD = 7

And, tan 60^o = CE/AE

√3 = CE/ 7

CE = 7√3 m

So, CD = CE + ED = (7√3 + 7)= 7(√3 + 1) m

Therefore, the height of the cable tower is 7(√3 + 1) m