Question

$A,B$ and $C$ are three points on a circle such that the angles subtended by the chords $AB$ and $AC$ at the centre $O$ are ${90}^{\circ }$ and ${110}^{\circ }$ respectively. Then the measure of angle $BAC$ is

• A. 75^o
• B. 80^o
• C. 50^o
• D. 40^o

Answer 1

The correct option is C: ${80}^{\circ }$

Given: $\angle BOA={90}^{\circ }$ and $\angle AOC={110}^{\circ }$

We know that angles around a point add up to ${360}^{\circ }$

$\therefore \angle BOC+\angle BOA+\angle AOC={360}^{\circ }$

$\Rightarrow \angle BOC+{90}^{\circ }+{110}^{\circ }={360}^{\circ }$

$\Rightarrow \angle BOC={360}^{\circ }-{200}^{\circ }$

$\angle BOC={160}^{\circ }$

We know that the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

Arc $BC$ is subtending $\angle BOC$ at the center and $\angle BAC$ on the remaining part of the circle, so $\angle BOC=2\times \angle BAC$

$\therefore \angle BAC=\frac{1}{2}\angle BOC$

$=\frac{1}{2}\times {160}^{\circ }$

$={80}^{\circ }$