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Question

A river is flowing with a speed of 1 kmh1. A swimmer wants to go to point C starting form A. He swims with a speed of 5 kmh1 at an angle θ with the river flow. If AB=BC=400 m, at what angle with the river bank should the swimmer swim?

A
θ=12sin12425
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B
θ=12sin11225
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C
θ=12sin11625
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D
θ=12sin1925
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Solution

The correct option is A θ=12sin12425
As AB = BC, swimmer has to travel equal distances in both directions. This implies that the resultant path of swimmer is at 45 with the river bank, therefore X- and Y- components of the swimmer's resultant must be equal.

Condition for reaching the point C is
tan45=VmYVmX
VmX=VmY


and we have, VmX=Vr+Vmcosθ
VmY=Vmsinθ

Therefore, to cover equal distances in both directions
Vr+Vmcosθ=Vmsinθ
1+5cosθ=5sinθ
5sinθ5cosθ=1
Squaring both sides
(sin2θ+cos2θ)sin2θ=125
sin2θ=2425
or θ=12sin12425

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