MCQ
In a parallelogram ABCD, if $\angle B A D=60^{\circ}$ and $\angle D B C=80^{\circ}$, then $\angle C D B=$
- A$140^{\circ}$
- B$20^{\circ}$
- ✓$40^{\circ}$
- D$120^{\circ}$
✓
Answer
Correct option: C.
$40^{\circ}$
(c) $40^{\circ}$
we have,
$\angle B A D=60^{\circ} \Rightarrow \angle B C D=60^{\circ}$
Using angle sum property in $\triangle D B C$, we obtain
$\angle D B C+\angle B C D+\angle B D C=180^{\circ} \Rightarrow 80^{\circ}+60^{\circ}+\angle B D C=180^{\circ} \Rightarrow \angle B D C=40^{\circ}$
we have,
$\angle B A D=60^{\circ} \Rightarrow \angle B C D=60^{\circ}$
Using angle sum property in $\triangle D B C$, we obtain
$\angle D B C+\angle B C D+\angle B D C=180^{\circ} \Rightarrow 80^{\circ}+60^{\circ}+\angle B D C=180^{\circ} \Rightarrow \angle B D C=40^{\circ}$
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