MCQ
In a $\triangle\text{ABC},$ if $\angle\text{A} = 60^\circ, \ \angle\text{B} = 80^\circ$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at $O$, the $\angle\text{BOC} =$
- ✓$120^\circ $
- B$150^\circ$
- C$30^\circ$
- D$60^\circ$
✓
Answer
Correct option: A.
$120^\circ $
$O$ is point where bisectors of $\angle\text{C}$ & $\angle\text{B}$ meets.
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$60^\circ + 80^\circ + \angle\text{C} = 180^\circ$
$\angle\text{C} = 40^\circ$
$\angle\text{C}2 = 20^\circ$
$\angle\text{C}2 = 20^\circ = \angle\text{BCO}\ ...\ (\text{i})$
$\angle\text{B}2 = 80^\circ = 40^\circ = \angle\text{OBC}\ ...\ (\text{ii})$
In $\triangle\text{BOC}$
$\angle\text{BCO} + \angle\text{OBC} + \angle\text{BOC} = 180^\circ$
From $(i)$ and $(ii)$
$20^\circ + 40^\circ + \angle\text{BOC} = 180^\circ$
$\angle\text{BOC} = 180^\circ - 60^\circ = 120^\circ$
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