Question
In the adjoining figure, $ABCD$ is a parallelogram in which $\angle\text{A}=70^{\circ}.$ Calculate $\angle\text{B},\angle\text{C}$ and $\angle\text{D}.$ 
✓
Answer
In a parallelogram, opposite angles are equal.
$\therefore\angle\text{A}=\angle\text{C}=72^{\circ}$
The sum of all the four angles of a parallelogram is $360^\circ$
So, $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^{\circ}$
$\Rightarrow72^{\circ}+\angle\text{B}+72^{\circ}+\angle\text{D}=360^{\circ}$
$[\because\angle\text{A} = \angle\text{C}]$
$\Rightarrow2\angle\text{B}+144^{\circ}=360^{\circ}$
$[\because\angle\text{B}=\angle\text{D}]$
$\Rightarrow2\angle\text{B}=360^{\circ}-144^{\circ}=216{^\circ}$
$\Rightarrow\angle\text{B}=\frac{216}{2}=108^{\circ}$
$\therefore\angle\text{B}=108^{\circ},\angle\text{C}=72^{\circ}$ and $\angle\text{D}=108^{\circ}.$
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