MCQ
In the given figure, $AOB$ is a diameter and $ABCD$ is a cyclic quadrilateral. If $\angle\text{ADC}=120^\circ,$ then $\angle\text{BAC}=?$
- A$20^\circ $
- B$45^\circ$
- C$60^\circ$
- ✓$30^\circ$
✓
Answer
Correct option: D.
$30^\circ$
We have:
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
$\Rightarrow\angle\text{ABC}+120^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=(180^\circ-120^\circ)=60^\circ$
$\Rightarrow\angle\text{ABC}=60^\circ$
Also, $\angle\text{ACB}=90^\circ$ (Angle in a semicircle)
In $\triangle\text{ABC},$ we have:
$\angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ (Angle sum property of a triangle)
$\Rightarrow\angle\text{BAC}+90^\circ+60^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=(180^\circ-150^\circ)=30^\circ$
$\Rightarrow\angle\text{BAC}=30^\circ$
$\angle\text{ABC}+\angle\text{ADC}=180^\circ$ (Opposite angles of a cyclic quadrilateral)
$\Rightarrow\angle\text{ABC}+120^\circ=180^\circ$
$\Rightarrow\angle\text{ABC}=(180^\circ-120^\circ)=60^\circ$
$\Rightarrow\angle\text{ABC}=60^\circ$
Also, $\angle\text{ACB}=90^\circ$ (Angle in a semicircle)
In $\triangle\text{ABC},$ we have:
$\angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ (Angle sum property of a triangle)
$\Rightarrow\angle\text{BAC}+90^\circ+60^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=(180^\circ-150^\circ)=30^\circ$
$\Rightarrow\angle\text{BAC}=30^\circ$
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