Question

In figure $A,B,C$ are three points on a circle such that the angles subtended by the chord $AB$ and $AC$ at the centre $O$ are ${80}^{\circ }$and ${120}^{\circ }$ respectively. Determine $\angle BAC$ and the degree measure of arc $BPC$.

Answer 1

In $△OAB,OB=OA=$ Radius.
Isosceles triangle, therefore, base angles are equal.
$\angle OAB=\angle OBA=\frac{({180}^{\circ }-{80}^{\circ })}{2}$ = ${50}^{\circ }$
Similarly in $△OAC,OC=OA=$ Radius Isosceles triangle, as base angles are equal.
$\therefore \angle OAC=\angle OCA=\frac{({180}^{\circ }-{120}^{\circ })}{2}={30}^{\circ }$
$\therefore \angle BAC=\angle BAO+\angle OAC={50}^{\circ }+{30}^{\circ }={80}^{\circ }$

$\Rightarrow \angle BOC={360}^{\circ }-({120}^{\circ }+{80}^{\circ })={160}^{\circ }$

Hence, length of arc $BPC=\frac{\theta }{{360}^{\circ }}\times 2\pi r$

$=\frac{{160}^{\circ }}{{360}^{\circ }}\times 2\pi r$

$=\frac{8}{9}\pi r$