- #1

APphysicsfail

- 1

- 0

## Homework Statement

Add these vectors: 10 km [S53°E] and 25 km 16.26° N of E

## Homework Equations

Law of cosines and law of sines equations

## The Attempt at a Solution

So I visualized the vectors as two arrows, the shorter one pointing down at a 323° angle (of 53° east of south) and the longer one starting at the tip of the first one, pointing up 16.26°. My first attempt was to split them up into components- I drew legs off of the lines to make right triangles and used trig to calculate Ax=3.96km, Ay=9.18km, Bx=21.29 and By=13.11 with Ax/Ay being the components of the 10km line, Bx/By the components of 25km. I then realized that the resultant line I was looking for wasn't going to be the hypotenuse of a triangle but the side of a quadrilateral, so finding the components was useless.

Then I found the supplementary angle between 37° (the compliment of 53°) and 16.26° to be 126.74°. I used the law of cosines to find the length of the resultant, with 126.74° being my angle C and the resultant being c.

c^2 = a^2+b^2 - 2ab(cosC)

c^2 = 10^2+25^2 - 2(10)(25)(cos126.74)

c^2 = 487.7

c = 22.08

I feel like I'm doing this wrong, since it's odd that the resultant is shorter than the second vector. I went on to plug this information into the law of sines to find the angle of the new vector, using 25 as B, 22.08 as C, 126.74 as c and looking for b.

sinb/B = sinc/C

sinb/25 = sin126.74/22.08

b = .665°

53°+0.665 = 53.665° E of S.

So the answer I got is 22.08km 53.665 E of S. I'm positive that this is wrong. Could someone please explain to me how this works and what I'm doing wrong? If my phrasing of the question is confusing, ask me any questions you have and I'll try convey it clearer. Thank you!