Question

Two chords $AB$ and $AC$ of a circle subtends angles equal to ${90}^o$ and ${150}^o$, respectively at the centre. Find $\angle BAC$, if $AB$ and $AC$ lie on the opposite sides of the centre.

• A. ${120}^o$
• B. ${240}^o$
• C. ${60}^o$
• D. ${30}^o$

Answer 1

The correct option is C ${60}^o$

Given- $AB$ and $AC$ are the chords of a circle with centre $O$.

$\angle AOB={90}^o$ and $\angle AOC={150}^o$.

To find out- $\angle BAC=?$

Solution-

$BC$ is joined.

Now $\angle BOC={360}^o-(\angle AOB+\angle AOC)={360}^o-({90}^o+{150}^o)={120}^o$.

So, $\angle BAC=\frac{1}{2}\angle BOC=\frac{1}{2}\times {120}^o={60}^o$ ....(The angle at the centre, subtended by a chord of a circle is double of that a the circumference.)