MCQ
In fig ABCD is a parallelogram. If $\angle\text{DAB}=60^\circ$ and $\angle\text{DBC}=80^\circ$ then $\angle\text{CDB}$ is:
- A80º
- B70º
- C40º
- D60º
✓
Answer
- 40º
Solution:
$40^\circ\ \angle\text{C} = 60^\circ$ as opposite angles of a parallelogram are equal and $\angle\text{CDB} = 40^\circ$ angle sum property of a triangle. [In $\triangle\text{CDB},\ \angle\text{C} + \angle\text{CDB} + \angle\text{DBC} = 180^\circ$]
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