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Q.

A river if flowing with a speed of 1 km/hr. A swimmer wants to go to point C starting from A. He swims with speed of 5 km/hr, at an w.r.t river flow. If AB = BC = 400 m. At what angle with river bank should swimmer swim 1. The value of $θ$

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a

52°

b

50°

c

53°

d

54°

answer is C.

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Detailed Solution

Question Image

$\begin{array}{l} Tan 45^{\circ} = \frac{v_{y}}{v_{x}} \Rightarrow v_{y} = v_{x} \\ v_{R} + v_{s} \cos θ = v_{s} \sin θ \\ 1 + 5 \cos θ = 5 \sin θ \end{array}$
On squaring on both sides
$\begin{array}{l} 1 + 25 \cos^{2} θ + 10 cor θ = 25 - 25 \cos^{2} θ \\ θ = 53^{\circ} \end{array}$

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