Question

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

Answer 1

Let $P_1$ and $P_2$ be the before and after positions of aeroplane

Given:

$P_1{G}_1=h=1\text{km}$

$\angle P_{1}O{G}_1={60}^{\circ }$

$\angle P_{2}O{G}_2={30}^{\circ }$

From $△O{P}_1{G}_1$

$\tan {60}^{\circ }=\frac{P_1{G}_1}{O{G}_1}\Rightarrow \sqrt{3}=\frac{1}{O{G}_1}$

$\Rightarrow O{G}_1=\frac{1}{\sqrt{3}}\text{km}$

From $△P_{2}O{G}_2$

$\tan {30}^{\circ }=\frac{P_2{G}_2}{O{G}_2}\Rightarrow \frac{1}{\sqrt{3}}=\frac{1}{O{G}_2}$

$\Rightarrow O{G}_2=\sqrt{3}\text{km}$

$G_1{G}_2=O{G}_2-O{G}_1=\sqrt{3}-\frac{1}{\sqrt{3}}=\frac{2}{\sqrt{3}}\text{km}$

$\text{Speed}=\frac{\text{Distance}}{\text{Time}}=\frac{\frac{2}{\sqrt{3}}}{\frac{10}{3600}}=240\sqrt{3}\text{km/hr}$