Question

A man in a boat rowing away from a lighthouse 150m high,takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60 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

Answer 2

Answer:Hey Dear,

Here is your answer

Step-by-step explanation:

Given :- root 3 = 1.732

Time taken = 2 minute

To find :- Speed of boat

Proof :- Distance travelled by boat in 2 minute = CD

In triangle ABC

P/B = AB / BC = tan 45°

150 / BC = 1 [ tan 45° = 1 ]

So, BC = 150m ----------(1)

In triangle ABD

P/B = AB / BD = tan 30°

150 / BC + CD = 1 / root 3 [ tan 30° = 1 / root 3 ]

150 / 150 + CD = 1 / root 3 [ from (1) ]

By cross multiplication

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 = 54.9 meter/minute

Hope this helps you

Have a nice day.