Question

An aeroplane flying horizontally 1 km above the ground is observed at an elevation of $60\degree$. If after 10 seconds, the elevation be $30\degree$, the uniform speed of the aeroplane is

• A. $240\sqrt3$ km/hr
• B. $\frac{240}{\sqrt3}$ km/h
• C. $\frac{120}{\sqrt3}$ km/hr
• D. $120\sqrt3$ km/h

Answer 1

Answer: (A)

$240\sqrt3$ km/hr

Answer 2

In $\triangle$ABC , tan 60° = $\frac{1}{BC}$
BC = $\frac{1}{\sqrt{3}}$
In $\triangle$DEC , tan 30° = $\frac{1}{EC}$
EC = $\sqrt{3}$
So , EB = d = EC − BC
d= $\sqrt{3}$ −$\frac{1}{\sqrt{3}}$ = $\frac{2}{\sqrt{3}}$
Now the speed of plane

$x$ = $\frac{\frac{2}{\sqrt{3}}}{\frac{10}{3600}}$ = 240$\sqrt{3}$ km/h.
So the correct option is (A).