Question

The angles of elevation of an airplane between two observation points, A and B, are 30 ° and 45 ° respectively. If the vertical height of the plane at the moment is 5 km, find the distance between A and B to the nearest tenth of a km.

Answer 1

xA = 10 km

xB = 7.071km

Explanation:

A/Q

p=5km

For point A

∅ = 30°

sin ∅ = p/h

sin 30° = 5/h

1/2 = 5/h

h = 5×2

h=10km

For point B

sin ∅ = p/h

sin 45° = 5/x

1/√2 = 5/x

x = 5 × √2

x = 7.071 km

.: Points A and B are 10 km and 7.071 km away from the airplane.