Question

a man in a boat rowing away from a lighthouse of 150 m high takes 2 minutes to change the angle of elevation of the top of the light house from 60° to 45°. find speed of 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

Answer 2

HYY MATE...✌️

Height of the Light house , p = 100 m

let initial distance be x m

and the angle is 60°

so, tan60° = p/x = 100/x

or, √3 = 100/x

or, x = 100/√3 m

Now after time 2min let the new distance be y

and angle is 45°

so, tan45° = 100/y

so, 1 = 100/y

so, y = 100 m

So, distance travelled in 2min

= y - x = 100 - 100/√3 = 100 - 57.737

= 42.263 m

So, d = 42.263 m

t = 2 min = 2×60 = 120 sec

speed = d/t = 42.263/120 m/s

= 0.35219 m/s

HOPE IT HELPS U....!!