# Magnitude of a hiker

1. Sep 3, 2009

### triplel777

1. The problem statement, all variables and given/known data

The route followed by a hiker consists of three displacement vectors A, B, and C. Vector A is along a measured trail and is 1550 m in a direction 28.0° 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 is 12.0° 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 vector B and vector C.

2. Relevant equations

3. The attempt at a solution

i drew the triangle. broke it into two equations with two unknowns in each.
x: 1550cos28+bcos311 + ccos168=0
y: 1550sin28+bsin311+csin168=0
which gave me b= -1138.78 and c= 635.99

Last edited: Sep 3, 2009
2. Sep 3, 2009

3. Sep 3, 2009

### Staff: Mentor

What are the three angles of your triangle? Hint: Try using the law of sines.

4. Sep 3, 2009

### triplel777

i drew the triangle. broke it into two equations with two unknowns in each.
x: 1550cos28+bcos311 + ccos168=0
y: 1550sin28+bsin311+csin168=0
which gave me b= -1138.78 and c= 635.99

and what about the angles they give in the problem?

5. Sep 3, 2009

### kuruman

Your two starting equations are correct, but you numbers are not. Redo the calculation. If you still get the same numbers, then you must show how you got them so that we can help you. Don't forget to set your calculator to "Degree" mode.

6. Sep 3, 2009

### Staff: Mentor

If you drew the closed triangle A-B-C and you figured out its angles, then you could use the law of sines to solve for the two unknown sides.

As kuruman said, your equations for the components are fine, but the solution is not.

What about them? Presumably you used them to find the angles that you used in your equations. You could also use them to figure out the angles in the triangle.