Question

The angle of elevation of an aeroplane from a point on the ground is ${45}^{\circ }$. After a flight of 15 seconds, the angle of elevation changes to ${30}^{\circ }$. If the aeroplane is flying at a height of 3000 m, then find the speed of the plane.

$(\text{Take}\sqrt{3}=1.732)$

• A. 150 km/h
• B. 146.4 km/h
• C. 146.4 m/s
• D. 150 m/s

Answer 1

The correct option is C

146.4 m/s

Let A be the point on the ground and the distance CE be the distance travelled by plane in 15 seconds. So,

$\text{In}\Delta ACB$

$tan{45}^{\circ }=\frac{BC}{AB}=\frac{3000}{x}$

$1=\frac{3000}{x}\therefore x=3000m$

$\text{In}\Delta ADE$

$tan{30}^{\circ }=\frac{DE}{AB}=\frac{3000}{y}$

$\frac{1}{\sqrt{3}}=\frac{3000}{y}y=3000\sqrt{3}=3000\left(1.732\right)=5196m$

$\text{Distance travelled = y - x}=5196-3000=2196m$

$\text{Speed}=\frac{\text{Distance}}{\text{time}}=\frac{2196}{15}=146.4m/s$