An aeroplane flying horizontally 1 km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°; find the uniform speed of the aeroplane in km per hour.

Let aeroplane be originally at point E and after 10 seconds it reaches point D.

In △ABE,

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

In △ACD,

$\text{tan 30°} = \dfrac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{CD}{AC} \\[1em] \Rightarrow AC = \sqrt{3}CD \\[1em] \Rightarrow AC = \sqrt{3} \times 1 = \sqrt{3} \text{ km}.$

From figure,

DE = BC.

BC = AC - AB = $\sqrt{3} - \dfrac{1}{\sqrt{3}}$

= $\dfrac{3 - 1}{\sqrt{3}}$

= $\dfrac{2}{1.732}$

= 1.1547 km.

∴ DE = 1.1547 km.

∴ Aeroplane travels 1.1547 km in 10 seconds.

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

Speed = $\dfrac{\text{Distance}}{\text{Time}} = \dfrac{1.1547}{\dfrac{1}{360}}$

= 1.1547 × 360

= 415.67 km/hr.

Hence, speed of aeroplane = 415.67 km/hr.