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 (in km) is:

1. $\Big(\dfrac{\sqrt{3}+1}{2}\Big)$

2. $\Big(\dfrac{3+\sqrt{3}}{2}\Big)$

3. $3+\sqrt{3}$

4. $\sqrt{3}+1$

Let the position of the aeroplane be A. Let h be the height of the aeroplane above the ground.

Let C and D be the positions of the two consecutive stones on the ground.

Let the distance from the closer stone C to the foot of the perpendicular B be x.

Then, CD = x + 1.

In △ABC,

$\Rightarrow \tan 60^{\circ} = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BC} \\[1em] \Rightarrow \sqrt3 = \dfrac{h}{x} \\[1em] \Rightarrow x = \dfrac{h}{\sqrt3} \text{ …..(1)}$

In △ABD,

$\Rightarrow \tan 45^{\circ} = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{BD} \\[1em] \Rightarrow 1 = \dfrac{h}{x + 1} \\[1em] \Rightarrow x + 1 = h \text{ …..(2)}$

Substituting value of x from equation (2) in (1), we get :

$\Rightarrow \dfrac{h}{\sqrt3} + 1 = h \\[1em] \Rightarrow h - \dfrac{h}{\sqrt{3}} = 1 \\[1em] \Rightarrow h \Big(1 - \dfrac{1}{\sqrt{3}} \Big) = 1 \\[1em] \Rightarrow h \Big( \dfrac{\sqrt{3} - 1}{\sqrt{3}} \Big) = 1 \\[1em] \Rightarrow h = \dfrac{\sqrt{3}}{\sqrt{3} - 1} \\[1em] \Rightarrow h = \dfrac{\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} \\[1em] \Rightarrow h = \dfrac{3 + \sqrt{3}}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow h = \dfrac{3 + \sqrt{3}}{3 - 1} \\[1em] \Rightarrow h = \dfrac{3 + \sqrt{3}}{2} \text{ km}$

Hence, option 2 is the correct option.