Question

An aeroplane flying at a height of 300 m above the ground passes vertically above another plane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. What is the height of the lower plane from the ground?

(a) 50 m

(b) \(\frac{100}{\sqrt3}\) m

(c) 100\(\sqrt3\) m

(d) 150 (\(\sqrt3\) + 1) m

Answer 1

 (c)  100\(\sqrt3\) m.

Let the positions of the two planes at the instant when they are vertically above each other be C and B respectively.

Let A be the base level on the ground from where the height of the upper plane is 300 m, i.e.,

AC = 300 m, AB = x metres (say) 

Let angle of elevations of P be the position from where the two planes are measured, i.e.,

∠CPA = 60°, ∠BPA = 45°,

PA = y m (say)

In rt. Δ ABP,

tan 45° = \(\frac{AB}{PA}\)

⇒ 1 = \(\frac{x}{y}\)

⇒ y = x ...(i)

In rt. Δ CPA,

tan 60° = \(\frac{AC}{PA}\)

⇒ \(\sqrt3=\frac{300}{y}\) 

⇒ y = \(\frac{300}{\sqrt3}\) ...(ii)

From (i) and (ii),

x = \(\frac{300}{\sqrt3}\)

= \(\frac{300}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\)

= 100\(\sqrt3\) m.

∴ The lower plane is flying at a height of 100\(\sqrt3\).