question_answer

The angle of elevation of a jet fighter from a point \[A\] on the ground is\[{{60}^{o}}\]. After a flight of\[10\,\,s\], the angle of elevation changes to\[{{30}^{o}}\]. If the jet is flying at a speed of\[432\,\,km/h\]. Find the constant height at which the jet is flying.

A) \[200\sqrt{3}m\]                             

B) \[400\sqrt{3}m\]

C) \[600\sqrt{3}m\]                             

D) \[800\sqrt{3}m\]

Correct Answer: C

Solution :

Since, speed of a flight be\[432\,\,km/h\]. \[\therefore \]Distance cover from \[A\] to\[B\],                 \[d=432\times \frac{5}{18}\times 10=1200\,\,m\] Now, in\[\Delta CBD\],                 \[\tan {{60}^{o}}=\frac{h}{x}\Rightarrow x=\frac{h}{\sqrt{3}}\] and in\[\Delta CAD\],                 \[\tan {{30}^{o}}=\frac{h}{d+x}\] \[\Rightarrow \]               \[\frac{1}{\sqrt{3}}=\frac{h}{1200+\frac{h}{\sqrt{3}}}\] \[\Rightarrow \]               \[1200=\sqrt{3}h-\frac{h}{\sqrt{3}}\] \[\Rightarrow \]               \[h=600\sqrt{3}m\]