Question

A man in a boat throwing away from a cliff 150 m high take 2 min to change the angle of elevation of the top of the cliff from 60 degree to 45 degree find the speed of the boat

Answer 1

Answer:

Height of cliff i.e. AB = 150 m

The angle of elevation ab  , ∠ACB = 6 °

Distance of boat from the bottom of the cliff = BC

Changed angle of elevation , ∠ADB = 45°

Now, Distance of boat from the bottom of the cliff = BD

Distance traveled by boat in 2 minutes = CD = BD-BC

In ΔABD

tanθ = Perpendicular / Base

tan45° =AB/BC

[∵tan45°=1]

1= 150/BC

BC =150/√3

So distance traveled by boat = 2 min

CD=BD-BC

150 + CD = 150 × root 3

150 + CD = 150 × 1.732 ( given )

150 + CD = 259.8

CD = 259.8 - 150

CD = 109.8 m

Speed = Distance / Time

Speed = 109.8 / 2

Speed of boat = 54.9 meter/minute

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@GauravSaxena01