Wow, tried helping a friend with a vector problem and I can't get it

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The problem involves a hiker returning to the trailhead after three legs of hiking, with the first leg being 1550m at 21 degrees north of east, the second leg at 41 degrees east of south, and the third at 18 degrees north of west. The discussion emphasizes that all vectors must sum to zero, leading to the establishment of equations based on the components of each leg. The calculations resulted in negative values for the distances, indicating an error in the approach or assumptions made about the angles. A suggestion is made to utilize the cosine and sine rules for better accuracy, and the importance of reality-checking results is highlighted, especially in multiple-choice scenarios. The conversation ultimately stresses the need for careful angle interpretation and verification of results in vector problems.
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Homework Statement



A hiker starts from the trailhead and hikes 1550m in a direction 21* north of east. Then, using her compass, she hikes the second leg of her journey in a direction 41* east of south. Finally, she hikes an additional distance in the direction 18* north of west. At the end of the third leg of her trip, she is back at the trailhead where she started. What is the distance she hiked along the second leg?

Homework Equations





The Attempt at a Solution



Ok.. so I took this as "all of the vectors sum to 0."

Converting all angles to "actual" angles, I have 21*, 311*, and 108*, respectively.

I then figured that I could make a system of equations, one equation for each coordinate so that they sum to 0. let n be the magnitude of the second vector, and q be the magnitude of the third.

ncos(311) + qcos(108) + 1550cos(21) = 0

and

nsin(108) + qsin(311) + 1550sin(21) = 0

Valid?

I then made everything into polynomials for simplicity and got these two equations:

.95q - .75n = -555
and
-.3q + .65n = -1447.04

Which shows that q = -3684 and that n = -3926.

So, n should be the magnitude of the second vector of the trip, right? My work doesn't lead to an answer choice.
 
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There are two standard approaches to this:
draw the resulting triangle and either;
1. resolve all vectors to components (put N along y-axis and E along x for eg); or
2. work out known interior angles and lengths: use the cosine and sine rules.

You know one length and all the angles so method #2 looks good for this one.

You tried #1 - treating "actual" angles as measured anticlockwise from due east.
But I think you want to take a closer look at that third angle 18 degrees north of West is 180-18 isn't it?

Apart from not leading to an answer choice - you should be able to see that the result is not even reasonable.
The second angle goes almost SE and only ends up 18 degrees below the start ... try to get into the habit of reality-checking your results.
Especially for multiple choice - because sneaky examiners will include common mistakes in the available choices.
 
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