Ship A is 10km due west of ship B. Ship A is heading directly north at a speed of 30 kmph while ship B is heading in a direction 60° west of north at a speed 20kmph. Their closest distance of approach will be........

Explanation

`vecV_(A) = 30 j, vecV_(B) = 20 sin 60^(@)hati + 20cos 60^(@)hatj`
`=-10sqrt(3i) + 10j`
`vecV_(BA) = vecV_(B)-vecV_(A) =-20hati -10sqrt(3)hatj`
If `phi` is the angle made by `vecV_(BA)` with X axis
`tan theta =20/(10sqrt(3)) =2/sqrt(3)`

and `sinphi =2/sqrt(7)`
From `triangleABC, x/10 = 2/sqrt(7)`
`x=20/sqrt(7) = 7.56` km