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°, what is the speed of the aircraft ? (tan = 15° = 0.2679)

Explanation

Let O be the ground observation point and A, B, C be the positions of the aircarft at t = 0s, t = 5s and t = 10 s respectively,

Clearly, `tan15^(@) = (AB)/(OB)` or `AB= OB tan 15^(@)`
Thus, distance travelled by the aircraft in 10s,
i.e., `s = AC = 2AB = 2(OB tan 15^(@))`
`= 2 xx 3400 xx 0.2679 = 1822 m`
Speed of the aircraft `= (s)/(t) = (1822m)/(10s) = 182.2 m//s`