Can some one help me on this algebra/geometry problem?
Its number 10 on the link. i can't post the question here because it includes a picture. Please show your work and answer.
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.