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$ABCD$ is a parallelogram in which $\angle A = 70^o$. Compute $\angle B, \angle C$ and $\angle D$.
Given:
$ABCD$ is a parallelogram in which $\angle A = 70^o$.
To do:
We have to compute $\angle B, \angle C$ and $\angle D$.
Solution:
We know that,
The opposite angles of a parallelogram are equal.
Adjacent angles of a parallelogram are supplementary
Therefore,
$\angle A+\angle B=180^o$
$70^o+\angle B=180^o$
$\angle B=180^o-70^o$
$\angle B=110^o$
$\angle C=\angle A=70^o$ (Opposite angles)
$\angle D=\angle B=110^o$ (Opposite angles)
Hence, $\angle B=110^o, \angle C=70^o$ and $\angle D=110^o$.
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