Simple Vector Sum Problem But My Calculated Angles Weird

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[SOLVED] Simple Vector Sum Problem... But My Calculated Angles Weird...

Homework Statement



A car travels 210 m [N32°E] and then travels 37 m [S51°W]. How far is the car from the origin?

Homework Equations



Cosine Law.
C^2=A^2+B^2-2(A)(B)(CosC)

Sine Law.
C/CosC=A/CosA

The Attempt at a Solution



C^2=210^2+37^2-(2)(210)(37)(Cos7)
C=173m

173/Sin7=37/Sin(theta)
Theta=1.5 Degrees

32 Degrees + 1.5 Degrees = [N33.5E]

Resultant solution = 173m [N33.5E]

I'm not sure if I got this question right... although I am sure that the magnitude is correct. However when I try to solve for all of the angles of the triangle, it does not add up to 180 Degrees. The problem is kind of funky because the angles are so small.
Can anyone tell me why the angles don't add up to 180 degrees... I have checked my solution 3 times but I still don't get where I could of made a mistake.

In fact, after I find out all the angles and add up I get this.
8.5 Degrees + 1.5 Degrees + 7 Degrees = 17 Degrees... That is way lower than 180.
I don't know how the hell I am getting 8.5 Degrees. I think it is because of some sine rule that I don't know, because 1.5 + 7 degrees = 8.5 degrees.
Can anyone explain this to me? Why is my calc saying 8.5 degrees instead of 171.5 Degrees.
 
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Seems like if you draw this out and add the vectors graphically, you get the one going north of east at 32 degrees, then coming from the head of that vector is the second smaller one going 51 degrees south of west, and the resultant vector goes from the tail of the first to the head of the second, right?

So in that triangle, seems like the angle between the two known vectors is 19 degrees, and from looking at the picture the resultant vector's angle(in terms of degrees north of east)should be LESS than the original, not more

Also looking at it graphically, you're going to have two small angles(one of which I believe is 19 degrees)and one big one larger than 90 degrees
 
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How did you get 19 degrees for the angle between the two known vectors.
So 90-51-32=19 eh? I would double check your subtraction.
I really don't think that is the problem at hand.
Look at my last sentence in the original post.
 
I'm just drawing the hell out of it and I still think it's 19(which is 51-32 fyi)

But let's say it was 7, I'm assuming you used the law of sines then?

Which you typed as C/CosC=A/CosA
and man I could've sworn the law of sines included a different trig function that wasn't COsine...

Not to mention that it looks like you used capital letters for the sides, though I think that part may've been a typo(so the angle across from side C is usually denoted c)
 
Can you explain your logic for subtracting 51-32? That whole quadrant is 90 degrees. We know that one of the angles is 32 degrees, due to opposite angles theorem. 90-32 is equal to 58 degrees, which is greater than 51 degrees. We know can conclude that there is a 7 degree angle between those two angles.

Wow you get my point. I am running on two cans of redbull here give me a break. Sine law would use Sine sorry for the mistake. And yes those capital letters to distinguish angles from length are typos.
 
At least the two cans of redbull is probably why you could make sense of my last paragraph there(I meant to point out you used caps for both, and the same one at that...)

As for the 19, I just drew it. I'd do this in paint and upload it except I'm lazy, soooo...draw both originating from the origin, with positive y-axis as north and all the usual stuff. So I have the larger vector going north east, and the other going southwest

You're trying to find the angle between them if you drew them like you were adding them, so if the tail of the second vector originates from the head of the first

So to find this angle, couldn't you extend the tail of the first vector down into quadrant three? It's a transverse line intersecting the x axis, so from geometry, the angle between the x-axis and the extension of vector 1 into quadrant 3 is 32 degrees, and we know that the angle between the x-axis and the second vector is 51 degrees and it's also in the third quadrant...
 
#@%@#%@#$I see my mistake now. Thanks. I drew [W51S] instead of [S51W].
God bless procrastination=hindersperformance.