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

In. Fig. O is the observation point at the ground. A and B are the positions of air craft for which `angleAOB=30^(@)`. Draw a perpendicular OC on AB. Here OC=3400m and `angleAOC=angleCOB=15^(@)`.
In `DeltaAOC, AC=OC tan45^(@) = 3400 xx 0.2679=910.86` m.
AB=AC+CB=AC+AC=2AC=`2xx 910.86`m
Speed of the aircraft, v=`("distance AB")/("time")= (2 xx 910.86)/(10)=182.17 ms^(-1)=182.2 ms^(-1)`.