Displacement Vectors?

Question

The route followed by a hiker consists of three displacement vectors, (vector)A, (vector)B and (vector)C. Vector A is along a measured trail and is 1550 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 41.0˚ east of south. Similarly, the direction of vector C is 35.0˚ north of west. The hiker ends up back where she started, so the resultant displacement is zero, or (vector)A+ (vector)B+ (vector)C =0. What are the magnitudes of vectors B and C?

So far I've come up with the Xcomponent of A being 1404.8 and the Ycomponent is 655.1. I also figured vector B to pointing at 319˚ (in reference to the coordinate plane) and vector C is pointing at 145˚.

Can anyone please give me any help possible?

Actually I was wrong, vector B is 229˚ on coordinate plane

Answer 1

|| A || = 1550m
∠A = 25.0˚
∠B = 270° + 41° = 311°
∠C = 180° - 35° = 145°

Sum of x-components:

1550cos 25.0˚ + Bcos 311° + Ccos 145° = 0

Sum of y-components:

1550sin 25.0˚ + Bsin 311° + Csin 145° = 0

So that's 2 equations in 2 unknowns:

Bcos 311° + Ccos 145° = -1550cos 25.0˚
Bsin 311° + Csin 145° = -1550sin 25.0˚

You can go ahead and get the approximations for the sines and cosines...but I think since it's gonna be messy anyway, I'd go for the determinant method:

| -1550cos 25.0˚ cos 145° |
| -1550sin 25.0˚ sin 145° |
- - - - - - - - - - - - - - - - - - - - =
| cos 311° cos 145° |
| sin 311°  sin 145°  |

-1550(cos 25.0˚ sin 145° - sin 25.0˚ cos 145°)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = 5548.6m = || B ||
(cos 311˚ sin 145° - sin 311˚ cos 145°)

| cos 311° -1550cos 25.0˚ |
| sin 311° -1550sin 25.0˚ |
- - - - - - - - - - - - - - - - - - - - =
| cos 311° cos 145° |
| sin 311° sin 145° |

-1550(cos 311˚ sin 25.0° - sin 311˚ cos 25.0°)
- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - = 6158.8m = || C ||
(cos 311˚ sin 145° - sin 311˚ cos 145°)

Answer 2

here x-axis is from west to east, while y-axis is from south to north;
A=1550(cos(25°)·i +sin(25°)·j);
B=|B|*(cos(-90°+41°)·i + sin(-90°+41°)·j);
C=|C|*(cos(180°-35°)·i + sin(180°-35°)·j);
Now A+B+C=0; or 1550(cos(25°)·i +sin(25°)·j) + |B|*(cos(-90°+41°)·i + sin(-90°+41°)·j) + |C|*(cos(180°-35°)·i + sin(180°-35°)·j)= 0·i +0·j;
Thus 2 equations with 2 unknown |B| and |C|, i.e.
■ +|B|*cos(49°) - |C|*cos(35°) =-1550*cos(25°);
■ -|B|*sin(49°) + |C|*sin(35°) =-1550*sin(25°);
matrix method for solving the system being preferable.
Check me though first!