UPSC CAPF Assistant Commandant Model Paper 6

GIVEN:
Angle of depression from an airplane to a point on the ground is 30°
After travelling for 300 km, angle of depression of same point = 60°
CONCEPT:
Height and distance
FORMULA USED:Tan θ = Perpendicular / Base
CALCULATION:

AB = 300 km
Let DE = x
∠BAE = ∠AEC = 30°
∠PBE = ∠BED = 60°
Let AC = BD = h
Now,
Tan 60° = BD / DE = h/x
⇒ √3 = h/x
⇒ h = x√3 ---- (1)
Tan 30° = AC / CE = h/(300 + x)
⇒ 1 / √3 = h / (300 + x)
⇒ h = (300 + x)/√3 ---- (2)
From (1) and (2):
(300 + x)/√3 = x√3
⇒ 300 + x = 3x
⇒ 2x = 300
⇒ x = 150
∴ Required distance travelled by airplane = x = 150 km