Find sin theta and tan theta?

Using

sin^2 ( theta) + cos^2 (theta)  = 1    we have

sin^2(theta) +  ( -12 / 13)^2  = 1

sin^(theta)  =  1 - (-12/13)^2

sin^2(theta)  =  1 - 144/169

sin^2(theta)  =    169/169  -  144/169

sin^2 (theta)  =  25/169       take the square root of both sides

sin (theta)  = ± √ [25/169]     and the sine is negative in the 3rd quadrant

sin (theta)  =  -5/13

tan (theta)  =   sin (theta) / cos (theta)  =  ( -5/13) / (-12/13)  =  5/12