heights and distances

(Pl see attached figure)

Let the poin5t of observation be A , initial position of the plane be P  and new position be Q . BP = CQ = 3000m

Angle PAB = 45 deg and Angle QAC = 30 deg

Consider triangle PBA . Since one angle is 90 deg and the other is 45deg , the third angle is also 45 deg , so AB = PB = 3000m

Now let BC = x . In triangle AQC  3000/ ( 3000+x) = tan 30 = 1/ 3

therefore , 30003 = 3000+x

3000(3 -1) = x

 

3000 x  0.732 = x

x= 2196m

speed = distance (in km)/ time (in hrs)

=2196 x 3600 / 15 x 1000 =  527 . 04 km / hr