# Q. 4.25 An aircraft is flying at a height of  $3400\; m$ above the ground. If the angle subtended at a ground observation point by the aircraft positions $10.0\; s$ apart is $30^{\circ},$ what is the speed of the aircraft?

The given situation is shown in the figure:-

For finding the speed of aircraft we just need to find the distance AC as we are given t = 10 sec.

Consider $\Delta$ ABD,

$\tan 15^{\circ}\ =\ \frac{AB}{BD}$

$AB\ =\ \ BD\ \times \tan 15^{\circ}$

or                                       $AC\ =\ 2AB\ =\ \ 2BD\ \times \tan 15^{\circ}$

or                                                  $=\ \ 2\times 3400\times \tan 15^{\circ}$

or                                                  $=\ 1822.4\ m$

Thus, speed of aircraft  :

$=\ \frac{1822.4}{10}\ =\ 182.24\ m/s$

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