question_answer

The angle of elevation of a fighter jet plane from a point on the ground is\[60{}^\circ \]. After 10s the angle of elevation becomes\[30{}^\circ \]. If the speed of jet plane is 432 km/h, then the height of the jet plane from the ground is

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 jet plane is 432 km/h \[\therefore \]Distance travelled from B to A, \[d=432\times \frac{5}{18}\times 10\] \[=1200m\] In\[\Delta CBD,\] \[\tan 60{}^\circ =\frac{h}{x}\] \[\Rightarrow \] \[x=\frac{h}{\sqrt{3}}\] and in\[\Delta CAD,\] \[\tan 30{}^\circ =\frac{h}{d+x}\] \[\Rightarrow \] \[\frac{1}{\sqrt{3}}=\frac{h}{1200+\frac{h}{\sqrt{3}}}\] \[\Rightarrow \] \[1200=\sqrt{3}h-\frac{h}{\sqrt{3}}\] \[\Rightarrow \] \[h=600\sqrt{3}m\]