An aircraft is flying at a constant height with a speed of 360 km/h. From a point on the ground, the angle of elevation of the aircraft at an instant was observed to be 45°. After 20 seconds, the angle of elevation was observed to be 30°. Determine the height at which the aircraft is flying (use $\sqrt{3}$ = 1.732).

Speed of aircraft = 360 km/h

Distance covered in 20 seconds = $\dfrac{360 \times 20}{60 \times 60} = 2$ km

Let aeroplane be flying at a hight of h km.

E is the fixed point on ground and A is the initial position of aircraft and C is the position after 20 seconds.

Considering right angled △EDC we get,

$\Rightarrow \text{tan 30°} = \dfrac{CD}{ED} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{h}{ED} \\[1em] \Rightarrow ED = h\sqrt{3}$

From figure,

$\Rightarrow ED = EB + BD \\[1em] \Rightarrow h\sqrt{3} = EB + 2 \\[1em] \Rightarrow EB = h\sqrt{3} - 2 \text{ …….(Eq 1)}$

Considering right angled △AEB we get,

$\Rightarrow \text{tan 45°} = \dfrac{AB}{EB} \\[1em] \Rightarrow 1 = \dfrac{h}{EB} \\[1em] \Rightarrow EB = h \text{ …….(Eq 2)}$

Comparing Eq 1 and Eq 2 we get,

$h = h\sqrt{3} - 2 \\[1em] \sqrt{3}h - h = 2 \\[1em] 0.732h = 2 \\[1em] h = \dfrac{2}{0.732} \\[1em] h = 2.732 \text{ km} = 2732 \text{ m}$

Hence, the aircraft is flying at a height of 2732 meters.