Question

Question 9
Two chords AB and AC of a circle subtends angles equal to ${90}^{\circ }and{150}^{\circ }$ respectively at the centre. Find $\angle BAC$, if AB and AC lie on the opposite sides of the centre.

Answer 1

In $\Delta BOA$
OB = OA [both are the radius of circle]
$\angle OAB=\angle OBA$ . . . . (i) [angles opposite to equal sides are equal]

In $\Delta OAB,\angle OBA+\angle OAB+\angle AOB={180}^{\circ }$ [by angle sum property of a triangle]
$\Rightarrow \angle OAB+\angle OAB+{90}^{\circ }={180}^{\circ }$ [from Eq (i)]
$\Rightarrow 2\angle OAB={180}^{\circ }-{90}^{\circ }$
$\Rightarrow 2\angle OAB={90}^{\circ }$
$\Rightarrow \angle OAB={45}^{\circ }$

Now, in $\Delta AOC$ AO = OC [both are the radius of a circle]
$\therefore \angle OCA=\angle OAC$ ............(ii) [angles opposite to equal sides are equal]
$Also,\angle AOC+\angle OAC+\angle OCA={180}^{\circ }$ [by angle sum property of a triangle]
$\Rightarrow {150}^{\circ }+2\angle OAC={180}^{\circ }[fromEq.(ii\left)\right]$
$\Rightarrow 2\angle OAC={180}^{\circ }-{150}^{\circ }$
$\Rightarrow 2\angle OAC={30}^{\circ }$
$\Rightarrow \angle OAC={15}^{\circ }$
$\therefore \angle BAC=\angle OAB+\angle OAC={45}^{\circ }+{15}^{\circ }={60}^{\circ }$