A river is flowing from west to east at a speed of 5 metre : A man on the south bank of the river capable of swimming at 10 metre per minute in still water , wants to swim across the river by the shortest path Calculate the direaction in which he should swin .

Explanation

The situation is shown in fig.
OA=A, represents velocity of river which is 5 `mt^(-1)`
OB = B, represents velocity of man in still water which is 10 m `m t^(-1)`
The man starts swimming from 0 and goes along OB tot reach C by shortest possible path.
Hence, `angle AOC = beta = 90^@ `
Here, A= 5 m `m t^(-1) , B = 10m mt^(-1) beta = 90^@ , theta = ?`
From formula,` tan beta = ( B sin theta )/(A+B cos theta )`
Putting values, we get, `tan 90^@ =(10 sin theta )/( 5+10 cos theta ) = oo ` (infinite)
Hence, `5+10 cos theta = 0 `or `10 cos theta =-5`
(or) `cos theta =-0.5 = cos 120^@ i.e., theta = 120^@ `
It makes `angle COB = 30^@ `
The man should swim along OB which is `30^@ ` west of north.