Question

$ABCD$ is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and if $\angle ADC={140}^o$, then $\angle BAC$ is equal to:

• A. ${60}^o$
• B. ${50}^o$
• C. ${40}^o$
• D. ${30}^o$

Answer 1

The correct option is B ${50}^o$

Given:

$ABCD$ is a cyclic quadrilateral with $AB$ as the diameter of the circle.

Also, $\angle ADC={140}^o.$

We have to find: $\angle BAC$.

Then,

$\angle ADC+\angle ABC={180}^o$ ....(since the sum of the opposite angles of a quadrilateral is ${180}^o$).

$\therefore \angle ABC={180}^o-{140}^o={40}^o.$

Also, $\angle ACB={90}^o$ ....(since the angle subtended by a diameter at the circumference of the circle, is ${90}^o$).

$\therefore$ In $\Delta ABC$ we have,

$\angle ACB={90}^o$ and $\angle ABC={40}^o.$

So, $\angle CAB={180}^o-(\angle ACB+\angle ABC)$

$={180}^o-({90}^o+{40}^o)$

$\therefore \angle BAC={50}^o$ .

Hence, option $B$ is correct.