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

The Correct Option is A

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).