Physics HelpPLease!?

The route followed by a hiker consists of three displacement vectors A,B,and C . VectorA is along a measured trail and is 2590 m in a direction 25.0 ° north of east. Vector B is not along a measured trail, but the hiker uses a compass and knows that the direction is 37.0 ° east of south. Similarly, the direction of vector C is 30.0 ° north of west. The hiker ends up back where she started, so the resultant displacement is zero, or A+B C + = 0. Find the magnitudes of (a) vector B (b) and vector C








I'm soo stuck and I know its easy. I tried adding by components and I got 2347 for Ax, .6018B for Bx, and -.866C for Cx. for Ay 1094, By -.7986 and .50C for Cy but they don't add up to zero so I got it wrong?? I also tried the law of sines but I got the answer wrong

Physics HelpPLease!?
First lets give Vector B length B and Vector C length C.


Then breaking down into components we have


2347.337 meters = C*cos(30)-B*37sin(37) (X components where X is along east west and Y is along North south)


1094.581 meters = B*cos(37)-C*sin(30)


Expand out the trig terms and you have:


2347.337 = .866C-.602B


1094.581=.799 B - .5 C





Solve system of equations for B and C gives:


C=1.597B-2189.163


2347.337 = 1.383B-1895.81-.602B


B=5430 m


C = 6484 m