A motor boat is racing towards North at 25 km/h and the water current has speed 10 km/h in the direction of `60^(@)` east of South. Find the resultant velocity of the boat.

Explanation

Refere to figure here

A=25km/h along OP,B=10km/h along oQ
Angle between `vecA` and` veB` is `theta=180^(@)-60^(@)=120^(@)`
`R=sqrt(A^(2)+B^(2)+2AB.cos theta)`
`=sqrt(25^(5)+10^(2)+2xx25xx10cos 120^(@))=21.8` km/h
`tan beta=(B sin theta)/(A+B cos theta)=(10xxsin 120^(@))/(25+10cos 120^(@))=0.433=tan 23.4^(@)`
or `beta=23.4^(2)` East of North