MCQ
In the given figure, $ABCD$ is a parallelogram in which $\angle\text{BDC}=45^{\circ}$ and $\angle\text{BAD}=75^{\circ}.$ Then, $\angle\text{CBD}=?$
- A$45^\circ$
- B$55^\circ$
- ✓$60^\circ$
- D$75^\circ$
✓
Answer
Correct option: C.
$60^\circ$
Since $ABCD$ is a parallelogram, $AB \| CD$ since opposite angles of a parallelogram are equal.
$\Rightarrow\angle\text{ABD}=\angle\text{BCD}=45^{\circ}$ ...(Alternate angles)
In $\triangle\text{ADB},$
$\angle\text{ABD}+\angle\text{BDA}+\angle\text{DAB}=180^{\circ}$ ...(Angle sum property)
$\Rightarrow45+\angle\text{BDA}+75=180$
$\Rightarrow\angle\text{BDA}+120=180$
$\Rightarrow\angle\text{BDA}=60^{\circ}$
$\Rightarrow\angle\text{CBD}=\angle\text{BDA}=60^{\circ}$ ...(Alternate angles)
$\Rightarrow\angle\text{CBD}=60^{\circ}$
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