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A man swims from a point A one bank of a river of width 100m. When he swims perpendicular to the water current, he reaches the other bank 50m downstream. The angle to the bank at which he should swim, to reach the directly opposite point B on the other bank is

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$10^{0} upstream$

$20^{0} upstream$

$30^{0} upstream$

$60^{0} upstream$

answer is D.

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Detailed Solution

$\tan θ = \frac{υ_{w}}{υ_{m}} = \frac{50}{100} = \frac{1}{2} o r υ_{m} = 2 υ_{w}$

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$\sin θ = \frac{υ_{w}}{υ_{m}} = \frac{υ_{w}}{2 υ_{w}} = \frac{1}{2} o r θ = 30^{o}$

So, it is $60^{o}$ upstream

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