A river is flowing with a speed of 1 km/hr. A swimmer wants to go to point $C$ starting from $A$ . He swims with a speed of 5 km/hr at an angle $θ$ with respect to the river. If $A B = BC = 400$ m, the value of $θ$ is:

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B is directly across from A and A B = BC , so the resultant velocity must point along A C at 4 5 ∘ : the across-river and along-river components are equal, 5 sin θ = 5 cos θ + 1 . Then 25 ( 1 − cos 2 θ ) = 25 cos 2 θ + 10 cos θ + 1 , giving 50 cos 2 θ + 10 cos θ − 24 = 0 , cos θ = 3/5 , θ = 5 3 ∘ . Option C.

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