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Three points $A, B, C$ are on the straight bank of a river such that $BC = 2$ km. $A$ boat is moving towards $B$ on a line at right angles to the bank. At Some point $AC$ subtends an angle of $60 ° .$ After moving for 10 minutes in the same direction $AC$ subtends an angle of $120 °$ . The speed of boat is

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4 \sqrt{3} km /

16 \sqrt{3} km /

8 \sqrt{3} km /

24 \sqrt{3} km /

answer is C.

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Detailed Solution

It is given that three points

A, B, C

are on the straight bank of a river and BC is 2km. After moving 10 minutes in the same direction

AC

subtends an angle of

120 °

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

EBC is right angle triangle and

∠ CEB

is half of

∠ AEC

.
So,

∠ AEC

60 °

\Rightarrow ∠ CEB = 30 °

Now, in triangle EBC,

∠ CEB = 30 °

∠ EBC = 90 °

then,

∠ ECB = 90 ° - 30 °

∠ ECB = 60 °

In triangle EBC,

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

\tan 60 ° = \frac{BE}{BC}

\Rightarrow \sqrt{3} = \frac{BE}{2}

∵ tan 60 ° = 3

∴ BE = 2 \sqrt{3}

km
Similarly, IBC is right angle triangle and

∠ BIC

is half of

∠ AIC

.
So,

∠ AIC

120 °

\Rightarrow ∠ BIC = 60 °

Now, in triangle IBC,

∠ BIC = 60 °

∠ IBC = 90 °

then,

∠ ICB = 90 ° - 60 °

∠ ICB = 30 °

In triangle IBC,

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

\tan 30 ° = \frac{IB}{BC}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{IB}{2}

∵ tan 30 ° = 13

\Rightarrow IB = \frac{2}{\sqrt{3}}

\Rightarrow IB = \frac{2 \sqrt{3}}{3}

km
For speed of boat,
Distance IE=EB-IB.
Substitute the values, we get

\Rightarrow IE = 2 \sqrt{3} - \frac{2 \sqrt{3}}{3}

\Rightarrow IE = \frac{4}{\sqrt{3}}

km
Now,

10

minutes

= \frac{10}{60} hr

=

\frac{1}{6}

hr
Speed will be,

\Rightarrow \frac{Distance}{Time}

\Rightarrow \frac{\frac{4}{\sqrt{3}}}{\frac{1}{6}}

\Rightarrow 8 \sqrt{3} km / hr

So the speed of the boat is

8 \sqrt{3} km / hr

.
So, the correct option is 3.

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