Question : $\mathrm{O}$ is the centre of a circle and $\mathrm{A}$ is a point on a major arc $\mathrm{BC}$ of the circle. $\angle \mathrm{BOC}$ and $\angle \mathrm{BAC}$ are the angles made by the minor arc $\mathrm{BC}$ on the centre and circumference, respectively. If $\angle \mathrm{ABO}=40^{\circ}$ and $\angle \mathrm{ACO}=30^{\circ}$, then find $\angle \mathrm{BOC}$.
Option 1: $130^{\circ}$
Option 2: $140^{\circ}$
Option 3: $120^{\circ}$
Option 4: $150^{\circ}$
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Correct Answer: $140^{\circ}$
Solution :
Given: That $O$ is the centre of a circle and $A$ is a point on a major arc $BC$ of the circle. $\angle BOC$ and $\angle BAC$ are the angles made by the minor arc $BC$ on the centre and circumference, respectively. Also $\angle ABO=40^{\circ}$ and $\angle ACO=30^{\circ}$.
We know the theorem states that in the circle, the angle subtended by an arc at the centre is twice the angle subtended by it at any point on the alternate segment.
$\angle BOC= 2 \times\angle BAC$
Again, $\angle BAC = \angle ABO + \angle ACO = 40^{\circ} + 30^{\circ} = 70^{\circ}$
So, $\angle BOC = 2 \times 70^{\circ} = 140^{\circ}$.
Hence, the correct answer is $140^{\circ}$.
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