Question

The angles of depression of the top and bottom of a building 50 metres high as observed from the top of a tower are 30° and 60°, respectively. Find the height of the tower and also the horizontal distance between the building and the tower.

Answer 1

In triangle BTP,

tan 30° = TP/BP

 

1/√3 = TP/BP

BP = TP√3

In ΔGTR,

tan 60° = TR/GR

√3 = TR/GR

GR = TR/√3

As BP = GR

TP√3 = TR/√3

3 TP = TP + PR

2 TP = BG

TP = 50/2 m = 25 m

Now, TR = TP + PR

TR = (25 + 50) m

Height of tower = TR = 75 m

Distance between building and tower = GR = TR/√3

GR = 75/√3 m = 25√3 m

Answer 2

Answer:

HeLlO

Here CD is the tower and AB is the pole.

In △CDB, CD/BC = tan 60°

=> 50m/BC = √3

=> BC = 50/√3m

AE = BC

AE = 50/√3m

In △ADE, DE/AE = tan 45°

=> DE/50/√3m = 1

=> DE = 50/√3m

so the height of the pole is 50/√3m