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A man on the top of a rock lying on a seashore observes a boat coming towards it. If it takes $10$ minutes for the angles of depression to change from $30^{\circ}$ to $60^{\circ}$ . How soon will the boat reach the shore?

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20

minutes

15

minutes

10

minutes

5

minutes

answer is B.

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Detailed Solution

It is given that a man on the top of a rock lying on a seashore observes a boat coming towards it. It takes

10

minutes for the angles of depression to change from

30^{\circ}

60^{\circ}

Let AB be the rock C and D be the two positions of the boat.
Question Image

∠ EAC = 30^{\circ}

and

∠ EAD = 60^{\circ}

As,

AE ∥ BC, ∠ ADB

and

∠ EAD

are alternate interior angles.
Now,

∠ ADB = ∠ EAD

⇒

∠ ADB = 60^{\circ}

As,

AE ∥ BC, ∠ EAC

and

∠ ACB

are alternate interior angles.

∠ EAC = ACB

⇒

∠ ACB = 30^{\circ}

Let

CD = x, DB = y

From

△ ABC

\tan 30^{0} = \frac{AB}{x + y}

⇒

AB = \frac{x + y}{\sqrt{3}}

From

△ ABD

\tan 60^{0} = \frac{AB}{y}

⇒

AB = \sqrt{3} y

⇒

\frac{x + y}{\sqrt{3}} = \sqrt{3} y

⇒

\frac{x}{2} = y

The boat takes

10

minutes to cover distance

x,

so it will take

5

minutes to cover

\frac{x}{2}

.
Time taken to reach

B

from point

C

= 10 + 5 = 15

minutes.

∴

Time in which boat reach the shore is

15

minutes.
Hence, the correct option is 2.

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A man on the top of a rock lying on a seashore observes a boat coming towards it. If it takes 10 minutes for the angles of depression to change from 30∘to 60∘. How soon will the boat reach the shore?

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A man on the top of a rock lying on a seashore observes a boat coming towards it. If it takes $10$ minutes for the angles of depression to change from $30^{\circ}$ to $60^{\circ}$ . How soon will the boat reach the shore?