A plane moves in windy weather due east while the pilot the plane somewhat south of east. The wind is blowing at 50 km/h directed `30^(@)` east of north, while the palne moves at 200 km/h relative to the wind . What is the velocity of the plane relative to the ground and what is the direction in which the pilot points the palne ?

Explanation

Strategy : Three vectors ` vecv _(PG), vec(PW) and vecv_(WG) ` are involved. We have to ifind the magnitude of ` vecv_(PG)` and the direction of ` vecv_(PW) (i.e. theta)` . Let us first relate the three vectors using the concept of relative velocity we can write.
( Velocity of plane w.r.t wind ) = ( veclocity of plane w.r.t ground ) - ( velocity of wind w.r.t ground )
` Rightarrow vecv_(PW) = vecv_(PG) - vecv_(WG)`
` Rightarrow vecv_(PG) = vecv_(PW) + vecv_(WG)`
Here we have two unknow quanitiies ` ( vecV_(PG) and theta)` , so we should not relate the amgnitudes of these vectors using triangle law. in such situations we resolve the vectors into components on the corrdinate system and then solve the equations for both axes ( x and y)
For the y comonents
` v_( "PG".y) = v_(PW.y) + V_(WG.y)`
i.e. 0= ` (-200) sintheta + ( 50 cos 3-0^(@))`
Solving we get ` sin theta = sqrt3/8 , theta = sin^(-1) (0.216) = 12^(@)`
For the x componnts
` v_(PG.y) = v_(PW.x) + v_(WG.x)`
`V_(PG) = 200 cos theta) + ( 50 sin 30^(@))`
` Rightarrow V_(PG) ` = 221 km/h