Find the resultant vector in situation 300km/h[45°]+400km⁄h[315°]?

300 [45°] + 400 [315°]
= ( 300*cos 45° i + 300*sin 45° j ) + ( 400*cos 315° i + 400*sin 315° j )
= ( 300*cos 45° i + 300*sin 45° j ) + ( 400*cos -45° i + 400*sin -45° j )
= ( 300*(√2 / 2 ) i + 300*(√2 / 2 ) j ) + ( 400*(√2 / 2 ) i + 400*(-√2 / 2 ) j )
= 150√2 i + 150√2 j + 200√2 i - 200√2 j
= 350√2 i - 50√2 j , 此向量在第四象限

tan θ = y / x = - 50√2 / 350√2 = - 1 / 7
θ = arctan ( - 1 / 7 ) ≒ 352°

r^2
= x^2 + y^2
= ( 350√2 )^2 + ( - 50√2 )^2
= 250000

r = √250000 = 500

Ans: 500 km/h [ 352° ]