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

Given:

$ABCD$ is a cyclic trapezium with $AD \| BC$.

$\angle B = 70^o$.

To do:

We have to determine other three angles of the trapezium.

Solution:

$AD \| BC$

This implies,

$\angle A + \angle B = 180^o$                    (Sum of the cointerior angles is $180^o$)

$\angle A + 70^o = 180^o$

$\angle A= 180^o- 70^o = 110^o$

$\angle A = 110^o$

Sum of opposite angles of a cyclic quadrilateral is $180^o$.

Therefore,

$\angle A + \angle C = 180^o$

$\angle B + \angle D = 180^o$

$110^o + \angle C = 180^o$

$\angle C = 180^o- 110^o = 70^o$

$70^o + \angle D = 180^o$

$\angle D = 180^o - 70^o = 110^o$

Hence,

$\angle A = 110^o, \angle C = 70^o$ and $\angle D = 110^o$.

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Updated on: 10-Oct-2022

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