Can some one help me on this algebra/geometry...

From the figure you can see that the side BA extended forms an
angle of 150 degree with line a. The line BC extended forms an angle of 118 degree with line
c.

Let the point at the end of BA extended be X, the point at the
end of BC extended be Y and the points at the right ends of the three parallel lines be A', B'
and C'.

So we have A'AX = 150 degrees. As a is parallel to b, the
angle B'BA is also equal  to 150 degrees.

Also, as C'CY is equal to
118 degrees and b is parallel to c we have B'BC = 118
degrees.

Therefore ABC = 360 - B'BA - B'BC = 360 - 150 - 118 = 92
degrees.

The magnitude of angle ABC is 92
degrees.