Question

In the figure $ABCD$, suppose $AB\parallel CD;\angle A={65}^{\circ }$ and $\angle B={75}^{\circ }$. What is the measure of $\angle D$ and $\angle C$?

Answer 1

Observe that. $\angle A+\angle D={180}^{\circ }$ (a pair of adjacent angles of a trapezium is supplementary). Thus ${65}^{\circ }+\angle D={180}^{\circ }$. This gives $\angle D={180}^{\circ }-{65}^{\circ }={115}^{\circ }$. Similarly $\angle B+\angle C={180}^{\circ }$, which gives ${75}^{\circ }+\angle C={180}^{\circ }$. Hence $\angle C={180}^{\circ }-{75}^{\circ }={105}^{\circ }$.