MCQ
In an isosceles $\triangle\text{ABC}$, the bisectors of $\angle\text{A}$ and $\angle\text{C}$ meet at a point $O$. If $\angle\text{A}$ = $40^\circ $, then $\angle\text{BOC}$ = ?
- ✓$110^\circ$
- B$70^\circ$
- C$130^\circ$
- D$150^\circ$
✓
Answer
Correct option: A.
$110^\circ$
In an isosceles $\triangle\text{ABC}$, $\angle\text{B}$ = $\angle\text{C}$ and bisector of $\angle\text{B}$ and $\angle\text{C}$ meet at $O$ and $\angle\text{A} = 40^\circ $
$\therefore\angle\text{B}=\angle\text{C}=\frac{\Big(180^\circ-40^\circ\Big)}{2}$
$=\frac{140^\circ}{2}=70^\circ$
$\therefore\frac{1}{2}\angle\text{B}=\frac{1}{2}\angle\text{C}=\frac{70^\circ}{2}=35^\circ$
Now in $\triangle\text{OBC}$
$\angle\text{BOC}+\angle\text{OBC}+\angle\text{OCB}=180^\circ$
$\angle\text{BOC}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{BOC}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-70^\circ=110^\circ$
$\therefore\angle\text{B}=\angle\text{C}=\frac{\Big(180^\circ-40^\circ\Big)}{2}$
$=\frac{140^\circ}{2}=70^\circ$
$\therefore\frac{1}{2}\angle\text{B}=\frac{1}{2}\angle\text{C}=\frac{70^\circ}{2}=35^\circ$
Now in $\triangle\text{OBC}$
$\angle\text{BOC}+\angle\text{OBC}+\angle\text{OCB}=180^\circ$
$\angle\text{BOC}+\frac{1}{2}\angle\text{B}+\frac{1}{2}\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{BOC}+35^\circ+35^\circ=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-70^\circ=110^\circ$
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