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Question

The angle of elevation of an aeroplane from a point on the ground is 45. After a flight of 15 seconds, the angle of elevation changes to 30. If the aeroplane is flying at a height of 3000 m, then find the speed of the plane.

(Take 3=1.732)


A

150 km/h

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B

146.4 km/h

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C

146.4 m/s

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D

150 m/s

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Solution

The correct option is

C

146.4 m/s



Let A be the point on the ground and the distance CE be the distance travelled by plane in 15 seconds. So,

In ΔACB

tan 45=BCAB=3000x

1=3000x
x=3000 m

In ΔADE

tan 30=DEAB=3000y

13=3000y
y=30003=3000(1.732)=5196 m

Distance travelled = y - x

=51963000=2196 m

Speed=Distancetime=219615=146.4 m/s


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