Mark M. answered 04/22/20
Retired Math prof with teaching and tutoring experience in trig.
Since B lies in quadrant 3, π < B < 3π/2. So, π/2 < B/2 < 3π/4. Therefore, cos(B/2) is negative since B/2 lies in the second quadrant.
cos(B/2) = -√[(1+cosB)/2] = -√(1/50) = -1/(5√2) = -√2/10