How do I find the x-component and y component for vector B?
You aren't supposed to post HW here; however, you can easily determine that by dissecting the second given angle into two smaller acute angles.
If you had to use "principal angles" and look sine and cosine up in a table you would reduce the angel. But a calculator will give the correct cosine and sine of 111 degrees.
Yes, <ux, uy>+ <vx, vy>= <ux+ vx,Uy+ Vy>. Find the x-component of each vector and add.
So how do I find theta 2? I drew the black lines in so it would be easier to identify the x and y components, but I dont know how to find the angle that theta 2 makes with the y-component of B
I misread what you had before. As I said, you are given that [itex]\theta_2= 111[/itex] but I was thinking that was also the angle with the horizontal. Since vector a makes 22 degrees with the horizontal itself, vector b makes an angle of 22+ 111= 133 degrees with the horizontal. The x-component of vector a is 15 cos(22) and the x component of vector b is 15 cos(133) (notice that the latter is negative). The x-component of their sum is the sum of those.
However, if you really want to find the angle vector b makes with the vertical (if it is another part of the problem), Just use 133- 90= 43. But be careful. If you try to use that angle for the first part of the problem, you will have to use sine rather than cosine and change the sign "by hand". Your calculator will tell you that cos(133)= -sin(43).
Thank you guys for the help.
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