MCQ
$ABCD$ is a cyclic quadrilateral such that $\angle\text{ADB}=30^\circ$ and $\angle\text{DCA}=80^\circ,$ then $\angle\text{DAB}=$
- A$125^\circ$
- ✓$70^\circ$
- C$150^\circ$
- D$100^\circ$
✓
Answer
Correct option: B.
$70^\circ$
It is given that $ABCD$ is cyclic quadrilateral $\angle\text{ADB}=90^\circ$ and $\angle\text{DCA}=80^\circ.$ We have to find $\angle\text{DAB}.$
We have the following figure regarding the given information
$\angle\text{BDA}=\angle\text{BCA}=30^\circ$ (Angle in the same segment are equal)
Now, since $ABCD$ is a cyclic quadrilateral
So, $\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\angle\text{DAB}+\angle\text{BCA}+\angle\text{DCA}=180^\circ\ [\angle\text{BCD}=\angle\text{BCA}+\angle\text{DCA}]$
$\angle\text{DAB}+30^\circ+80^\circ=180^\circ$
$\angle\text{DAB}=180^\circ-110^\circ$
$\angle\text{DAB}=70^\circ$
We have the following figure regarding the given information
$\angle\text{BDA}=\angle\text{BCA}=30^\circ$ (Angle in the same segment are equal)
Now, since $ABCD$ is a cyclic quadrilateral
So, $\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\angle\text{DAB}+\angle\text{BCA}+\angle\text{DCA}=180^\circ\ [\angle\text{BCD}=\angle\text{BCA}+\angle\text{DCA}]$
$\angle\text{DAB}+30^\circ+80^\circ=180^\circ$
$\angle\text{DAB}=180^\circ-110^\circ$
$\angle\text{DAB}=70^\circ$
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