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 s` apart is ` 30^@`, what is the speed of the aircraft ?

Explanation

Let `O` be observation point on the ground, and `A,B,C` be the positions of the aircraft at `t=0s,t=5s` and `t=10s` respectively.
Clearly, `tan15^(@)=(AB)/(OB)` or `AB=OB tan15^(@)`
Thus, distance travelled by the aircraft in `10s,AC=2AB=2(OB tan 15^(@))=2xx3400xx0.2679=1822m` speed of the aircraft
`v=(1822m)/(10s)=182.2m//s`.
Position vector: Position vector of point `A` with respect to `O`