In the figure, $ABCD$ is a parallelogram in which $\angle DAB = 75^o$ and $\angle DBC = 60^o$. Compute $\angle CDB$ and $\angle ADB$.
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Given:
$ABCD$ is a parallelogram in which $\angle DAB = 75^o$ and $\angle DBC = 60^o$.
To do:
We have to compute $\angle CDB$ and $\angle ADB$.
Solution:
We know that,
The opposite angles of a parallelogram are equal.
Adjacent angles of a parallelogram are supplementary
Therefore,
$\angle A+\angle B=180^o$
$75^o+\angle B=180^o$
$\angle B = 180^o - 75^o$
$\angle B = 105^o$
Therefore,
$\angle DBA = 105^o -60^o = 45^o$
$\angle CDB = \angle DBA = 45^o$ (Alternate angles)
$\angle ADB = \angle DBC = 60^o$ (Alternate angles)
Hence, $\angle CDB=45^o$ and $\angle ADB=60^o$.
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