Question

ABCD is a cyclic quadrilateral in a circle with centre O. If $angleADC={130}^o$ Find $\angle BAC$

• A. ${80}^o$
• B. ${40}^o$
• C. ${120}^o$
• D. ${134}^o$

Answer 1

The correct option is B

${40}^o$

ABCD is a cyclic quadrilateral and $\angle ADC={130}^o$ O is the centre of the circle, AOB is the diameter.

$\angle ABC={180}^o-{130}^o={50}^o$

In triangle ABC, $\angle BAC+\angle CBA={90}^o$

$\angle BAC+{50}^o={90}^o$

$\angle BAC={90}^o-{50}^o={40}^o$