Magnitude of Hiker's Vectors A, B & C

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SUMMARY

The hiker's displacement problem involves three vectors: A, B, and C, with vector A measuring 1550 m at 28.0° north of east. The hiker's vector B is directed 41.0° east of south, while vector C is 12.0° north of west. The resultant displacement is zero, leading to the equations: 1550cos(28) + bcos(311) + ccos(168) = 0 and 1550sin(28) + bsin(311) + csin(168) = 0. The calculated magnitudes for vectors B and C are approximately -1138.78 m and 635.99 m, respectively, indicating a need for recalculation and verification of angles used.

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Homework Statement



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.


Homework Equations





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:
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Show some work so that we can help you along.
 
What are the three angles of your triangle? Hint: Try using the law of sines.
 
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?
 
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.
 
triplel777 said:
i drew the triangle.
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.

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

and what about the angles they give in the problem?
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.
 

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