MCQ
$\text{ABCD}$ is a cyclic quadrilateral such that $\angle\text{ADB} = 30^\circ$ and $\angle\text{DCA} = 80^\circ,$ then $\angle\text{DAB} =$
- ✓$70^\circ$
- B$100^\circ$
- C$125^\circ$
- D$150^\circ$
✓
Answer
Correct option: A.
$70^\circ$
$\text{ABCD}$ is a cyclic Quadrilateral.
Consider $\triangle\text{ABD}$ and $\triangle\text{ABC}.$
Both are on the same base $AB$ and $\angle\text{ADB}$ and $\angle\text{ACB}$ are the angles in the same segment $AB$.
$\Rightarrow\angle\text{ADB}=\angle\text{ACB}=30^\circ$
$\Rightarrow\angle\text{BCD}=80^\circ+30^\circ=110^\circ$
In a cyclic Quadrilateral, sum of opposite angles is $180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\Rightarrow\angle\text{DAB}=180^\circ-110^\circ=70^\circ$
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