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.
tan‌15°=

AC

OC

or AC=OC‌tan‌15° =3,400×0.2679 =910.86m
Since the distance AC is covered by the aircraft in 5 s, the speed of the aircraft,
v=

910.86

5

=182.2ms−1