Question
A quadrilateral $ABCD$ is inscribed in a circle such that $AB$ is a diameter and $\angle\text{ADC}=130^\circ.$ Find $\angle\text{BAC}.$
✓
Answer
Draw a quadrilateral $ABCD$ inscribed in a circle having centre $O.$ Given,
$\angle\text{ADC}=130^\circ$ Since, ABCD is a quadrilateral. inscribed in a circle,
therefore $ABCD$ becomes a cyclic quadrilateral.
$\because$ Since, the sum of opposite angle of a cyclic quadrilateral is $180^\circ .$
$\therefore\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow130^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=50^\circ$
Since, $AB$ is a diameter of a circle, then $AB$ subtends an angle to the circle is right angle.
$\therefore\angle\text{ACB}=90^\circ$ In $\triangle\text{ABC},\ \ \angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ [by angle sum property of a triangle]
$\Rightarrow\angle\text{BAC}+90^\circ+50^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-(90^\circ+50^\circ)$
$=180^\circ-140^\circ=40^\circ$
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