Question

ABCD is a cyclic trapezium with AD || BC. If $\angle B={70}^{\circ }$, determine other three angles of the trapezium.

Answer 1

In the figure, ABCD is a trapezium in which AD || BC and $\angle B={70}^o$.

$\because$ AD || BC

$\therefore \angle A+\angle B={180}^o$ (Sum of cointerior angles)

$\Rightarrow \angle A+{70}^o={180}^o\Rightarrow \angle A={180}^{\circ }-{70}^{\circ }={110}^{\circ }$

$\therefore \angle A={110}^{\circ }$

But $\angle A+\angle C={180}^{\circ }and\angle B+\angle D={180}^o$

(Sum of opposite angles of a cyclic quadriteral)

$\therefore {110}^o+\angle C={180}^o$

$\Rightarrow \angle C={180}^o-{110}^o={70}^o$

and ${70}^o+\angle D={180}^o\Rightarrow \angle D={180}^o-{70}^o={110}^o$

$\therefore \angle A={110}^o,\angle C={70}^{o}and\angle D={110}^o$