Question
ABC is a triangle in which $\angle\text{A}=72^\circ,$ the internal bisectors of angles B and C meet in O. Find the magnitude of $\angle\text{BOC}.$
✓
Answer
Given, ABC is a triangle where $\angle\text{A}=72^\circ$ and the internal bisector of angles B and C meeting O. In $\triangle\text{ABC},$$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$$\Rightarrow72^\circ+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{B}+\angle\text{C}=180^\circ-72^\circ$
Dividing bith side by '2'$\Rightarrow\frac{\angle\text{B}}{2}+\frac{\angle\text{C}}{2}=\frac{108^\circ}{2}$
$\Rightarrow\angle\text{OBC}+\angle\text{OCB}=54^\circ$
Now, In $\triangle\text{BOC}\Rightarrow\angle\text{OBC}+\angle\text{OCB}+\angle\text{BOC}=180^\circ$$\Rightarrow540^\circ+\angle\text{BOC}=180^\circ$
$\Rightarrow\angle\text{BOC}=180^\circ-54^\circ=126^\circ$
$\therefore\angle\text{BOC}=126^\circ.$
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