Question

From an airplane vertically above a straight horizontal road, the angles of depression of two consecutive kilometers stones on opposite sides of the airplane are observed to be ${60}^o$ and ${30}^o$. Show that the height of airplane above the road is $\frac{\sqrt{3}}{4}$ km.

Answer 1

The angle of depression towards the roads on opposite side are ${30}^{\circ }$ and ${60}^{\circ }.$

Let x and y be the distance on the opposite side of an aeroplane. Let 'h' be the height of aeroplane from the road.

$x+y=1$ (given consecutive mile markers) -----(1)

$\tan {30}^{\circ }=\frac{h}{x}=\frac{1}{\sqrt{3}}$

$⟹x=\sqrt{3}h$

$\tan {60}^{\circ }=\frac{h}{y}=\sqrt{3}$

$y=\frac{h}{\sqrt{3}}$

putting these values in (1), we get

$⟹\sqrt{3}h+\frac{h}{\sqrt{3}}=1$

$⟹\frac{3h+h}{\sqrt{3}}=1$

$⟹\frac{4h}{\sqrt{3}}=1$

$⟹h=\frac{\sqrt{3}}{4}km$