Question

Two chords $AB$ and $AC$ of a circle subtends angles equal to ${90}^{\circ }$ and ${150}^{\circ }$ respectively at the centre. Find $\angle BAC$.

Answer 1

In $\Delta BOA$,
$OB=OA$ [radii]
$\angle OAB=\angle OBA$ ... (i) [angles opposite to equal sides are equal]

In $\Delta OAB$,
$\angle OBA+\angle OAB+\angle AOB={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow \angle OAB+\angle OAB+{90}^{\circ }={180}^{\circ }$
$\Rightarrow 2\angle OAB={180}^{\circ }-{90}^{\circ }$
$\Rightarrow \angle OAB={45}^{\circ }$
Now, in $\Delta AOC$,
$AO=OC$ [radii]
$\therefore \angle OCA=\angle OAC$ ... (ii) [angles opposite to equal sides are equal]
Also,
$\angle AOC+\angle OAC+\angle OCA={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow {150}^{\circ }+2\angle OAC={180}^{\circ }$ [from (ii)]
$\Rightarrow 2\angle OAC={180}^{\circ }-{150}^{\circ }$
$\Rightarrow \angle OAC={15}^{\circ }$
$\therefore \angle BAC=\angle OAB+\angle OAC={45}^{\circ }+{15}^{\circ }={60}^{\circ }$