Question

In the trapezium $\text{ABCD}$ ,side $\text{AB}$ is parallel to side $\text{DC}$ .If $\angle \text{A}={78}^{\circ }\text{and}\angle \text{C}={120}^{\circ }$, find angle $\text{B and D}$

Answer 1

Given Trapezium $\text{ABCD};$
$\angle \text{A}={78}^{\circ }\text{and}\angle \text{C}={120}^{\circ };\text{AB}||\text{CD}$


Thus ,
$\angle \text{(A+D)}={180}^{\circ }$ ..(1) [co-interior angle ]
Similarly
$\angle \text{(B+C)}={180}^{\circ }$ ..(2)
By using equation(1) and (2) and value of $\angle \text{A}$
${78}^{\circ }+\angle \text{D}={180}^{\circ }$($\therefore \angle \text{A}={78}^{\circ })$
$\therefore \angle \text{D}={102}^{\circ }$

By using equation (2) and value of $\angle \text{C}$
$\angle \text{B}+{120}^{\circ }={180}^{\circ }$$(\angle \text{A}={120}^{\circ })$
$\therefore \angle \text{B}={60}^{\circ }$
Hence ,value of $\angle \text{B and D}$ is ${60}^{\circ }\text{and}{120}^{\circ }$ respectively.