< is angle sign in this case.

C is the resultant of the forces, 2Aand 3B, that isC=2A+3B.

IfA=300N<43 deg. andC=100N<97 deg. , use the component method to determine the magnitude and direction ofB.

I am not sure if this problem is done correctly. Here is my attempt:

Ax = 600 cos 43 = 438N

Ay = 600 sin 43 = 409 N

Bx = ?

By = ?

Cx=100 sin 43 = 68 N

Cy=100 cos 43 = 73 N

438 + Bx= 68

409 + By = 73

Therefore

Bx= -370N

By= -336N

B = sqrt( (-370)^2 + (-336)^2) SoB =499N

Is this correct? What about the 3B?