1. ## System of equations

could someone please help with how youd want to solve this system? I have not taken a linear algebra course and have never dealt with a system like this.

3. Originally Posted by pickslides

A cos 210 + 200 cos Theta = 250 Cos 135 - 200 Cos 65
A sin 210 + 200 sin Theta = 250 sin 135 - 200 sin 65

4. Originally Posted by ur5pointos2slo
A cos 210 + 200 cos Theta = 250 Cos 135 - 200 Cos 65
A sin 210 + 200 sin Theta = 250 sin 135 - 200 sin 65
The first thing I would do is go ahead and find those sines and cosines.
Assuming that the arguments are degrees:
cos(210)= cos(180+ 30)= -cos(30)= $\displaystyle -\sqrt{3}/2$.
sin(210)= sin(180+ 30)= sin(30)= $\displaystyle \frac{1}{2}$
cos(135)= cos(90+ 45)= -cos(45)= $\displaystyle -\sqrt{2}/2$
sin(135)= sin(90+ 45)= sin(45)= $\displaystyle \sqrt{2}/2$
cos(65) and sin(65) are the only ones that cannot be written in a simple form. Are you sure it wasn't sin(60) and cos(60)?

5. Originally Posted by HallsofIvy
The first thing I would do is go ahead and find those sines and cosines.
Assuming that the arguments are degrees:
cos(210)= cos(180+ 30)= -cos(30)= $\displaystyle -\sqrt{3}/2$.
sin(210)= sin(180+ 30)= sin(30)= $\displaystyle \frac{1}{2}$
cos(135)= cos(90+ 45)= -cos(45)= $\displaystyle -\sqrt{2}/2$
sin(135)= sin(90+ 45)= sin(45)= $\displaystyle \sqrt{2}/2$
cos(65) and sin(65) are the only ones that cannot be written in a simple form. Are you sure it wasn't sin(60) and cos(60)?
Yes sir 100% sure it is 65 degrees.

6. ok so now that I have found the cosines and sines..How would I go about solving this thing?