MCQ
- A75º
- B55º
- C60º
- D45º
✓
Answer
- 60º
Solution:
As per the question
$\angle\text{BAD} = \angle\text{BCD} = 75^\circ$ (opposite angles of parallelogram)
Now, in $\triangle\text{BCD},$
$\angle\text{BCD} + \angle\text{CBD} + \angle\text{BCD} = 180^\circ$
$45^\circ + \angle\text{CBD} + 75^\circ = 180^\circ$
$\angle\text{CBD} = 60^\circ$
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