Q11. The angle of elevation of a jet plane from a point A on the ground is 60^o. After 4 flight of 15 seconds, the angle of elevation changes to 30^o. If the jet plane is flying at a constant height of 1500√3 meter, find the speed of the jet plane. (√3 = 1.732)

Solution :

Height of the plane from the ground PM = RN = 1500√3 m.

Angle of elevation are 30^o and 60^o.

From the figure

tan 60° =

PM

QM

√ =

√

QM

QM =

√

√

= m
Also tan 30° =

RN

QN

1

√

=

√

QM + MN

QM + MN = √ × √

+ MN =

MN = -

MN = mts.

∴ Distance travelled in 15 seconds = mts.

∴Speed of the jet plane =

distance

time

=

15

= m/s
= ×

18

5

kmph = kmph

Speed = m/sec. or kmph.