Question

The angles of depression of the top and bottom of a building 50 meters 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

AD=Tower
BC= Building
AB=Horizontal dis. b/w them
ang. APD=30°
ang. BPC=60°
H. OF Tower=50m
Hori. dis. b/w Them=200√3/3m

Answer 2

In ΔBTP,

tan 30° = TP/BP

[ See the attached figure ]

So,

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 = TR/√3

GR = 75/√3

m = 25√3 m