your method is basically correct, however it would be the result
If A + B = C
You have 2A + 3B = C
so rework 2Ax + 3Bx = Cx
etc
< is angle sign in this case.
C is the resultant of the forces, 2A and 3B, that is C=2A+3B.
If A=300N<43 deg. and C =100N<97 deg. , use the component method to determine the magnitude and direction of B.
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) So B = 499N
Is this correct? What about the 3B?
Ok so I did notice one error in my attempt above..This is my second attempt. Does it look correct? Can someone please work it out and let me know
Ax = 600 cos 43 = 438N
Ay= 600 sin 43 = 409 N
Bx=?
By=?
Cx=100 sin 7 = 12N
Cy=100 cos 7 = 99N
438 + 2Bx = 12
409 + 2By= 99
Bx=-213
By=-155
B = 263 N
Theta = 216 deg. from due east(+)