A motor boat is racing towards North at `25 km//h` and the water current in that region is `10 km//h` in the direction of `60 ^(00` East of South. Find the resultant velocity of the boat.

Explanation

Let O be the origin of journey of the motor boat
Here, `vecv_(b)`=velocity of motor boat
`=25km//hr`, due north
`vecv_(w)` =velocity of water current
`=10km//hr, 60^(@)` east of south
`theta=120^(@)`
The magnitude of the resultant velocity `|vecv_(R)|` is given by:
`|vecv_(R)|=sqrt(v_(b)^(2)+v_(w)^(2)+2v_(b)+v_(w)cos120^(@))`
`=sqrt(25^(2)+10^(2)+2xx25xx10xx((-1)/(2)))`
`=21.8 km//hr`
Let `vecv_(R)` make an nagle of `beta` with `vecv_(b)` . Then
`tan beta=(v_(w)sintheta)/(v_(b)+v_(w) cos theta)=(10xxsin 120^(@))/(25+10 cos 120^(@))`
`=((10xxsqrt3)/(2))/(25-10xx(1)/(2))=(sqrt3)/(4)`
`beta=tan^(-1)((sqrt3)/(4))=tan^(-1)(0.433)=23.4^(@)`