Question
In figure, $ABCD$ is a parallelogram in which $\angle\text{DAB}=75^\circ$ and $\angle\text{DBC}=60^\circ$.
Compute $\angle\text{CDB},$ and $\angle\text{ADB}$.
Compute $\angle\text{CDB},$ and $\angle\text{ADB}$.
✓
Answer
To find $\angle\text{CDB}$ and $\angle\text{ADB}$
$\angle\text{CBD}=\angle\text{ABD}=60^\circ$ [Alternate interior angle. $AD∥ BC$ and $BD$ is the transversal] in $\angle\text{BDC}$
$\angle\text{CBD}+\angle\text{C}+\angle\text{CDB}=180^\circ$ [Angle sum property]
$\Rightarrow60^\circ+75^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\angle=180^\circ-(60^\circ+75^\circ)$
$\Rightarrow\angle\text{CDB}=45^\circ$
Hence, $\angle\text{CDB}=45^\circ,\angle\text{ADB}=60^\circ$
$\angle\text{CBD}=\angle\text{ABD}=60^\circ$ [Alternate interior angle. $AD∥ BC$ and $BD$ is the transversal] in $\angle\text{BDC}$
$\angle\text{CBD}+\angle\text{C}+\angle\text{CDB}=180^\circ$ [Angle sum property]
$\Rightarrow60^\circ+75^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\angle=180^\circ-(60^\circ+75^\circ)$
$\Rightarrow\angle\text{CDB}=45^\circ$
Hence, $\angle\text{CDB}=45^\circ,\angle\text{ADB}=60^\circ$
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