3340.12 – Trapezoid Angle Measure

The figure below, not drawn to scale, shows a trapezoid, $ABCD$ with $AB \parallel CD$ . The diagonal $BD$ is drawn and $BD =AD, \angle DCB = 110^\circ$ , and $\angle CBD = 30^\circ$ . What is $\angle ADB$ ?

Solution

Using co-interior angles, $110^\circ + 30^\circ + \angle B1 = 180^\circ$ . Then, $\angle D1 = B1 = 40^\circ$ and $\angle A = 40^\circ$ .

Therefore $\angle D2 = 100^\circ$

That's the answer: 100 degrees.