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A man on the top of a rock observed a boat coming towards the rock with a uniform speed. It takes $15$ minutes for the angle of depression to change from $60^{\circ}$ to $30^{\circ}$ , then in what time will the boat take to reach the shore?

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5

minutes

10

minutes

7 \frac{1}{2}

minutes

2 \frac{1}{2}

minutes

answer is C.

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Detailed Solution

It is given that a man on the top of a rock observed a boat coming towards the rock with a uniform speed.
It takes

15

minutes for the angle of depression to change from

60^{\circ}

30^{\circ}

Let, the height of the rock be

h

so,

\tan 30^{o} = \frac{h}{BD}

⇒

BD = h \sqrt{3}

\tan 6 0^{o} = \frac{h}{BC}

⇒

BC = \frac{h}{\sqrt{3}}

Now, the distance travelled by boat in 15 minutes is

CD

CD = BD - BC

⇒

CD = h \sqrt{3} - \frac{h}{\sqrt{3}}

Let

v

be the uniform speed of the boat.
⇒

v = \frac{h \sqrt{3} - \frac{h}{\sqrt{3}}}{15}

⇒

v = \frac{2 h}{15 \sqrt{3}}

Now to reach the shore it has to travel distance

BC = \frac{h}{\sqrt{3}}

in time

t

.
⇒

t = \frac{\frac{h}{\sqrt{3}}}{\frac{2 h}{15 \sqrt{3}}}

⇒

t = 7 \frac{1}{2}

minutes
Hence, the correct option is 3.

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A man on the top of a rock observed a boat coming towards the rock with a uniform speed. It takes 15 minutes for the angle of depression to change from 60∘to 30∘, then in what time will the boat take to reach the shore?

Start JEE / NEET / Foundation preparation at rupees 99/day !!

Detailed Solution

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A man on the top of a rock observed a boat coming towards the rock with a uniform speed. It takes $15$ minutes for the angle of depression to change from $60^{\circ}$ to $30^{\circ}$ , then in what time will the boat take to reach the shore?