Question

The angles of elevation of an aeroplane flying vertically above the ground as observed from two consecutive stones 1 km apart are 45° and 60° .The height of the aeroplane above the ground ok km is .​

Answer 1

Answer:

Two consecutive kilometre stones ⇒ C and D

∠ADB = 45°; ∠ACB = 60°

CD = 1 km.

AB = height of plane = h metre

BC = x metre (let)

In ∆ABC,

tan60° = AB/BC

√3 = h/x

h = √3x metre -----i

In ∆ABD

tan45° = AB/BD

1 = h/x + 1

h = x + 1

h = h/√3 + 1

from equation 1, we get

h - h/√3 = 1

√3 h - h/√3 = 1

h = √3 /√3 - 1

h = √3 ( √3 + 1)/( √3 - 1) ( √3 + 1)

h = √3 ( √3 + 1)/2 metre