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a man from the top of a light house sees a boat with an angle of depression of 30 degrees. it takes 10 minutes for the angle of depression to change from 30 degrees to 60 degrees.in what time does the boat reach from that point to the light house?

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let 'h ' be the height of the light house
Let the velocity of the boat (per min) be ‘v’
Let the time taken to reach the shore from angle of depression as 60 be ‘t’
Then the distance from it’s position to the shore is vt (distance=speed x time)
Let A be the top of the light house
D the bottom of the light house(or the shore)
C be the point where angle of depression is 60
B be the point where angle of depression is 30
Distance covered in 10 minutes=10v
Distance covered in t minutes =vt
In triangle ACD
Tan 60=h/vt
√3=h/vt
vt√3=h
multiplying with √3 on both sides
3vt=h√3…..(1)
In triangle ABD
Tan 30 = h/10v + vt
1/√3=h/10v+vt
(cross multiplying)
10v +vt=h√3……..(2)
Equating (1) and (2)
10v + vt=3vt
10v=2vt
10=2t
t=5 minutes
10 years agoHelpfull: Yes(1) No(0)

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