MCQ
- A80º
- B-60º
- C100º
- D60º
✓
Answer
- 60º
Solution:
Since ABCD is a cyclic quadrilateral
$\angle\text{B}+\angle\text{D}=180^\circ$
$60^\circ+\angle\text{D}=180^\circ$
$\angle\text{D}=120^\circ$
Now since AD is parallel to BC
$\angle\text{C}+\angle\text{D}=180^\circ$
$\angle\text{C}+120^\circ=180^\circ$
$\angle\text{C}=60^\circ$
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