Question

A man in a boat rowing away from a lighthouse 100m high , takes 2 mins to change the ∠ of elevation ot the top ofthe lighthouse from 60° to 45° . Find the speed of the boat.

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Answer 1

Step-by-step explanation:

Step-by-step explanation:

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

Going away from the light house with a speed of 0.35219 m/s.

HOPE IT HELPS YOU ❤

Answer 2

Answer:

tan 60=height /intial distance from lighthouse

=

root 3=100/x

=x=100/root 3 =initial distance from lighthouse =57.7m

then final distance from lighthouse similarly will be

tan45= height/y

y=final dist from lighthouse

y=100

so dist traveled = y-x =100-57.7=

=42.3

and time taken= 120 secs

so speed = dist. / time

=42.3/120= 0.35m/s

hope it helps

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