I have:
$A=-R_2\sin{t_0}$
$B=R_2\cos{t_0}$
$C=R_1\cos{t_0}$
$D=R_1\sin{t_0}$
So on the right side I have three variables ($R_1, R_2, t_0$) but on the left-side I have four variables. Is it correct to assume that it should be possible to have three variables on the left side as well? If so, how?
You can use substitution to eliminate one of the variables. From last equation you can extract $R_1=D/\sin t_0$. By plugging the result on the third one $$A=-R_2\sin{t_0}\\ B=R_2\cos{t_0}\\ D=C\tan t_0$$ you get three variables on the left $\{A,B,D\}$ and three on the right $\{R_2,t_0,C\}$