Suppose that O is the observation point on the ground. The aircraft is flying along XY at a height OC = 3,400 m from the ground. Let A and B be two positions of the aircraft 10 s apart. Thus, the aircraft goes from the point A to C (or from the point C to B) in 5 s. If the angle subtended by AB is 30° at the point O, then the angle subtended by AC (distance covered in 5 s) at O is 15° From the right angled ΔOAC. $\tan 15^{\circ}= \frac{AC}{OC}$ or $AC=OC \tan 15^{\circ}$ $=3{,}400 \times 0.2679$ $=910.86\,\text{m}$Since the distance AC is covered by the aircraft in 5 s, the speed of the aircraft,$v= \frac{910.86}{5}=182.2\,\mathrm{m\,s^{-1}}$