question_answer

The angle of elevation of an aeroplane from a point on the ground is \[{{45}^{o}}\]. After a flight of 10 sec, the elevation changes to \[{{30}^{o}}\]. If the aeroplane is flying at a height of 3 km, find the speed of the aeroplane.

A)  \[785.46\text{ }km/hr\]          

B)  \[790.56\text{ }km/hr\]                      

C)  \[780.56\text{ }km/hr\]          

D)  \[782.65\text{ }km/hr\]          

Correct Answer: B

Solution :

Let the point on the ground is E which is y metres from point B and let after 10 sec. flight it covers x metres distance. In  \[\Delta \,\,AEB,\]                         \[\tan \,{{45}^{o}}=\frac{AB}{EB}\,\Rightarrow 1=\frac{3000}{y}\Rightarrow \,y=3000\,m\]                                                     ?...(1) In \[\Delta CED,\] \[\tan \,{{30}^{o}}=\frac{CD}{ED}\Rightarrow \frac{1}{\sqrt{3}}=\frac{3000}{x+y}\]                                     \[(AB\,=\,CD)\] \[\Rightarrow \]   \[x+y=3000\sqrt{3}\]                        .....(2) From equation (1) and (2), we have             \[x+3000=3000\sqrt{3}\] \[\Rightarrow \]            \[x=2196\,m\] Speed of aeroplane \[=\frac{Distance\text{ }covered}{Time\text{ }taken}\] \[=\frac{2196}{10}\times \frac{18}{5}km/hr=790.56km/hr\] Hence, the speed of aeroplane is \[790.56\text{ }km/hr.\]