Question

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

• A. ${90}^{\circ }$
• B. ${60}^{\circ }$
• C. ${120}^{\circ }$
• D. ${80}^{\circ }$

Answer 1

The correct option is B

${60}^{\circ }$

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