In triangle ABC, AC = BC, angle DCB is 40 degrees, and CD is parallel to AB. What is the number of degrees in angle ACB?
If DCB = 40°......then BC is a transversal cutting parallels CD and AB
So angle CBA = angle DCB
And since AC = BC, then angle CAB = angle CBA = 40°
So...angle ACB = 180 - 2(40) = 180 - 80 = 100°
