The angle of elevation of a jet plane from a point A on the ground...

$\tan 60 ° = \dfrac{h}{y}$

$\sqrt{3} = \dfrac{h}{y} ⇒ h = \sqrt{3}y \ldots \left(1\right)$

$\tan 30 ° = \dfrac{h}{x + y}$

$\dfrac{1}{\sqrt{3}} = \dfrac{h}{x + y} ⇒ \sqrt{3}h = x + y \ldots \left(2\right)$

Speed $432km{h}^{- 1} ⇒ \dfrac{432 \times 20}{60 \times 60} ⇒ \dfrac{12}{5}km$

$\sqrt{3}h = \dfrac{12}{5} + y$

$\sqrt{3}h - \dfrac{12}{5} = y$

from \left(1\right)$</p><p>$h = \sqrt{3}\left[\sqrt{3}h - \dfrac{12}{5}\right]$</p><p>$h = 3h - \dfrac{12\sqrt{3}}{5}$</p><p>$h = \dfrac{6\sqrt{3}}{5}km$</p><p>$h = 1200\sqrt{3}m