cos t = − 12 /13 , terminal point of t is in Quadrant III

Its -4

cos t = − 12 /13     take the inverse cos....so we have

arccos [-12/13]  = t = about 157.4°

But...since this angle lies in Quad 3, its value will be   =   180 + [180 - 157.4]  ≈  202.6°