An aeroplane flying horizontally 1 km above the ground is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°. Find the speed of the aeroplane in km/h.

Let initially aeroplane be at point B and after 10 seconds it is at point C.

Since, aeroplane is flying horizontally 1 km above the ground so, BE = CD = 1 km.

Considering right angled triangle △ABE,

$\Rightarrow \text{tan 60°} = \dfrac{BE}{AE} \\[1em] \Rightarrow \sqrt{3} = \dfrac{1}{AE} \\[1em] \Rightarrow AE = \dfrac{1}{\sqrt{3}} \\[1em] \Rightarrow AE = 0.557 \text{ km}.$

From figure,

AD = AE + ED = (0.577 + ED) km.

Considering right angled triangle △ACD,

$\Rightarrow \text{tan 30°} = \dfrac{CD}{AD} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{1}{0.577 + ED} \\[1em] \Rightarrow 0.577 + ED = \sqrt{3} \\[1em] \Rightarrow ED = 1.732 - 0.577 \\[1em] \Rightarrow ED = 1.155 \text{ km}.$

Aeroplane covers 1.155 km in 10 seconds.

Time = 10 seconds = $\dfrac{10}{60 \times 60} = \dfrac{10}{3600} = \dfrac{1}{360}$ hours.

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{1.155}{\dfrac{1}{360}} = 1.155 \times 360 =$ 415.66 km/h.

Hence, the speed of aeroplane is 415.66 km/h.