a swimmer can swim in still water at the rate of 4 kilometre per hour he was to choose a river flowing along a straight course at the rate of 2 km per hour so as to reach directly the opposite point on the other bank in what direction should we attempt to swim
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so these are vectors. the red one is our velocity 40, blue is water's velocity 20 and the yellow one is the velocity we want, to be able to cover shortest distance. that's why it's vertically upward. we need to find the angle theta so the the yellow resultant is vertically upwards.
from trigonometry, we know that
cos(theta) = base/hypotenuse.
so here, cos(theta) = 20/40 = 0.5.
this implies
theta = arccos (0.5) = 60°.
arccos is just basically the inverse of cos. it means that cos(60°) will be 0.5 so arccos(0.5) is 60°.
that's the answer. we should swim at an angle of 60° to the horizontal.
Hope it helped :)
from trigonometry, we know that
cos(theta) = base/hypotenuse.
so here, cos(theta) = 20/40 = 0.5.
this implies
theta = arccos (0.5) = 60°.
arccos is just basically the inverse of cos. it means that cos(60°) will be 0.5 so arccos(0.5) is 60°.
that's the answer. we should swim at an angle of 60° to the horizontal.
Hope it helped :)
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