Here, `4km//h` is the velocity of rain with respect to the man.
`v_(r//m)=4km//h`, vertically downward
`v_(m//g)=3 km//h` towards west
`v_(r//g)=?`
Since `v_(r//m)` is resultant of `v_(r//g)` and `v_(m//g)` (in opposite direction)
`u:v_(r//g)` (actual velocity of rain)
`tan 90^(@)=(u sin(90+theta))/(3+ucos(90+theta))`
`1/0=(ucos theta)/(3-usin theta)implies u sin theta=3.......(i)`
`[cos(90+ theta)=-sin theta]`
`(v_(r//m))^(2)=(v_(m//g))^(2)+(v_(r//g))^(2)+2(v_(m//g))(v_(r//g)) cos (90+ theta)`
`(4)^(2)=(3)^(2)+u^(2)+2(3)(u)(-sin theta)`
`16=9+u^(2)-2xx3xx3 implies u^(2)=25 implies uu=5km//h`
`usin theta=3 implies 5 sin theta=3implies sin theta=3/5 implies theta=37^(@)`
Rain is falling with a speed of `5km//h` at an angle `37^(@)` with vertical.
OR
Component approach
`u sin theta-3 =0 implies u sin theta=3`
`u cos theta=4`
` u=sqrt((usin theta)^(2)+(u cos theta)^(2))=5km//h`
`tan theta=(u sin theta)/(u cos theta)=3/4 implies theta=37^(@)`
OR
`vec(v)_(r//m)=-4hatk,vec(v)_(m//g)=-3hatj`
`vec(v)_(r//m)=vec(v)_(r//g)-vec(v)_(m//g)`
`=-4hatk=vec(v)_(r//g)-(-3hatj)`
`vec(v)_(r//g)=-3hatj-4hatk`
`|vec(v)_(r//g)|=v_(r//g)=sqrt((-3)^(2)+(-4)^(2))=5km//h`
`tan theta=3/4 implies 37^(@)`
