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$75^\circ $
- B$55^\circ$
- ✓$60^\circ$
- D$45^\circ$
✓
Answer
Correct option: C.
$60^\circ$
As per the question
$\angle\text{BAD} = \angle\text{BCD} = 75^\circ$ (opposite angles of parallelogram)
Now, in $\triangle\text{BCD},$
$\angle\text{BCD} + \angle\text{CBD} + \angle\text{BCD} = 180^\circ$
$45^\circ + \angle\text{CBD} + 75^\circ = 180^\circ$
$\angle\text{CBD} = 60^\circ$
$\angle\text{BAD} = \angle\text{BCD} = 75^\circ$ (opposite angles of parallelogram)
Now, in $\triangle\text{BCD},$
$\angle\text{BCD} + \angle\text{CBD} + \angle\text{BCD} = 180^\circ$
$45^\circ + \angle\text{CBD} + 75^\circ = 180^\circ$
$\angle\text{CBD} = 60^\circ$
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