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$ABCD$ 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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