Question
In the adjoining figure, ABCD is a parallelogram in which $\angle\text{DAB}=80^{\circ}$ and $\angle\text{DBC}=60^{\circ}.$ Calculate $\angle\text{CDB}$ and $\angle\text{ADB}.$
✓
Answer
ABCD is a parallelogram,
so opposite angles are equal.
$\therefore\angle\text{C}=\angle\text{A}=80^{\circ}$
As AD || BC and BD is a transversal.
So, $\angle\text{ADB}=\angle\text{DBC}=60^{\circ}$ [Alternate angles]
In $\triangle\text{ABD}$
$\angle\text{A}+\angle\text{ADB}+\angle\text{ABD}=180^{\circ}$
$\Rightarrow80^{\circ}+60^{\circ}+\angle\text{ABD}=180^{\circ}$
$\Rightarrow140^{\circ}+\angle\text{ABD}=180^{\circ}$
$\Rightarrow\angle\text{ABD}=180^{\circ}-140^{\circ}=40^{\circ}$
$\therefore\angle\text{ABC}=\angle\text{ABD}+\angle\text{DBC}$
$=40^{\circ}+60^{\circ}=100^{\circ}$
In a parallelogarm, opposite angles are equal.
So, $\angle\text{ADC}=\angle\text{ABC}=\angle100^{\circ}$
$\therefore\angle\text{CDB}=\angle\text{ADC}-\angle\text{ADB}$
$=100^{\circ}-60^{\circ}=40^{\circ}$
and $\angle\text{ADB}=60^{\circ}.$
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