Question

The angle of elevation of a jet plane from a point A on the ground is 60°. After a flight of 15 seconds, the angle of elevation changes to 30°. If the jet plane is flying at a constant height of $1500\sqrt{3}\mathrm{m}$, find the speed of the jet plane.

Answer 1

Let C and C' be the two positions of the jet plane and BC and B'C' be its constant heights.
Thus, we have:
$$\begin{gathered} \angle CAB=60°\mathrm{and}\angle C\text{'}AB\text{'}=30° \\ \mathrm{Let}: \\ AB=x\mathrm{and}BB\text{'}=y \end{gathered}$$



Now, in the right ∆ABC, we have:
$$\begin{gathered} \tan 60°=\frac{BC}{AB} \\ \Rightarrow \sqrt{3}=\frac{1500\sqrt{3}}{x} \\ \Rightarrow x=1500 \end{gathered}$$

Now, in the right ∆AB'C', we have:
$$\begin{gathered} \tan 30°=\frac{B\text{'}C\text{'}}{AB\text{'}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{1500\sqrt{3}}{x+y} \\ \Rightarrow x+y=1500\times 3=4500 \\ \Rightarrow y=4500-1500 \\ \Rightarrow y=3000 \end{gathered}$$

​BB' is 3000 m. This means that to cover a distance of 3000 m, the jet plane takes 15 seconds.
∴ Speed of the jet plane $=\frac{3000\mathrm{m}}{15\sec }=200\raisebox{1ex}{$\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\sec $}\right.=200\times \frac{18}{5}=720\raisebox{1ex}{$\mathrm{km}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{h}$}\right.$