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A man on a cliff observes a boat at an angle of depression $30 °$ which is sailing towards the shore to the point immediately beneath him. $3$ minutes later the angle of depression of the boat is found to be $60 °$ . Assuming that the boat sails at a uniform speed, the time taken by the boat to reach the shore is

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1.5

min

4.5

min

6

min

6.5

min

answer is B.

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Detailed Solution

It is given a man on a cliff observes a boat at an angle of depression is

30 °

.The boat is sailing towards the shore to the point beneath him. After

3

minutes the angle of depression of the boat is

60 °

.
According to the given data, the figure has been drawn below.
Question Image

Let

AB

be the cliff, speed be

s

, total time taken be

T

, total distance BD be

X

, height of cliff is

h

and

∠ BAC = 30 °

and

∠ BAD = 60 ° .

We know

speed = \frac{distance}{time}

\Rightarrow s = \frac{X}{T}

\Rightarrow X = sT

Let the distance covered in

3

min is

x = 3 s

.
In the figure CD represents the distance

x

.
From

ΔABC

, we get

tan θ = P e r p e n d i c u l a r b a s e

\tan 30 ° = \frac{BC}{AB}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{BC}{AB}

∵ tan 30 ° = 13

\Rightarrow AB = BC \sqrt{3} . . . . . . (1)

From the figure,

BD = BC + CD

\Rightarrow BC = BD - CD

= X - x

….(2)
Now equation 1 becomes,

\Rightarrow AB = \sqrt{3} (X - x) . . . . . . (3)

From

ΔABD

, we get

tan θ = P e r p e n d i c u l a r b a s e

\tan 60 ° = \frac{BD}{AB}

\Rightarrow \sqrt{3} = \frac{X}{AB}

∵ tan 60 ° = 3

\Rightarrow AB = \frac{X}{\sqrt{3}} . . . . . . . . . . . . . (4)

From

(3), (4)

\Rightarrow \sqrt{3} (X - x) = \frac{X}{\sqrt{3}}

\Rightarrow X = 3 (X - x)

\Rightarrow X = 3 X - 3 (3 s)

……{

x = 3 s

}

\Rightarrow 3 X - X = 9 s

\Rightarrow 2 X = 9

\Rightarrow 2 (sT) = 9 s

….{

X = sT

}

\Rightarrow

2 T = 9

\Rightarrow T = \frac{9}{2}

= 4.5

min
So, the time required to reach the shore is 4.5 min.
Hence, the correct option is 2.

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