The angle of elevation of an aircraft from a point A on the ground is $\displaystyle 60$°. After a flight of $\displaystyle 30$ seconds, the angle of elevation changes to $\displaystyle 30^{\circ}$. The aircraft is flying at a constant height of $\displaystyle 3500 \sqrt{3} \mathrm{~m}$ at a uniform speed. Find the speed of the aircraft.

Let P and Q be the positions of aircraft at two different times. \[\begin{aligned} & \tan 60^{\circ}=\sqrt{3}=\frac{3500 \sqrt{3}}{x} \\ & \Rightarrow x=3500 \mathrm{~m} \ldots \text { (i) } \\ & \tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{3500 \sqrt{3}}{x+y} \Rightarrow x+y=10500 \end{aligned} \] Using (i), we get \(\displaystyle \mathrm{y}=7000\) ∴ Speed of aircraft \(\displaystyle =\frac{7000}{30}=\frac{700}{3}\) or $\displaystyle 233.3$ m/s approx.