Question

In the given isosceles trapezium $ABCD,\angle C={102}^{\circ }.$ Find all the remaining angles of the trapezium.

Answer 1

From the figure, it is given that, Isosceles trapezium $ABCD$, $\angle C={102}^{\circ }$

$AB\parallel CD$

We know that sum of adjacent angles is equal to ${180}^{\circ }$

$So,\angle B+\angle C={180}^{\circ }$

$\begin{array}{c}\angle B+{102}^{\circ }={180}^{\circ }\\ \angle B={180}^{\circ }-{102}^{\circ }\\ \angle B={78}^{\circ }\end{array}$

Then, $AD=BC$

$\begin{array}{c}So,\angle A=\angle B\\ {78}^{\circ }={78}^{\circ }\end{array}$

Sum of all interior angles of trapezium is equal to ${360}^{\circ }.$

$\begin{array}{c}\angle A+\angle B+\angle C+\angle D={360}^{\circ }\\ {78}^{\circ }+{78}^{\circ }+{102}^{\circ }+\angle D={360}^{\circ }\\ 258+\angle D={360}^{\circ }\end{array}$

By transposing we get,

$\begin{array}{c}\angle D={360}^{\circ }-{258}^{\circ }\\ \angle D={102}^{\circ }\end{array}$