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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Updated on: 10-Oct-2022

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