In ABC, A=60^ , B=80^ , and the bisectors of B and C, meet at O. … - Vidyadip

Question

In $\angle \text{ABC}, \angle \text{A}=60^\circ,\ \angle \text{B}=80^\circ, $ and the bisectors of $\angle \text{B}$ and $\angle \text{C},$ meet at $O$. Find.
$i. \angle \text{C}$
$ii. \angle \text{BOC}$

✓

Answer

We know that the sum of all three angles of a triangle is $180^\circ $
Hence, for $\triangle \text{ABC},$ we can say that:
$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ ($Sum of angles of $\triangle \text{ABC})$
$60^\circ+80^\circ+\angle \text{C}=180^\circ$
$\angle \text{C}=180^\circ-140^\circ$
$\angle \text{C}=40^\circ$
For $\angle \text{OBC},$
$\angle \text{OBC}=\angle \text{B}2=\frac{80}{2} (OB$ bisects $\angle \text{B})$
$\angle \text{OBC}=40^\circ$
$\angle \text{OCB}=\angle \text{C}2=\frac{40}{2} (OC$ bisects $\angle \text{C})$
$\angle \text{OCB}=20^\circ$
If we apply the above logic to this triangle, we can say that:
$\angle \text{OCB}+\angle \text{OBC}+\angle \text{BOC}=180^\circ$ $($Sum of angles of $\triangle \text{OBC})$
$20^\circ+40^\circ+\angle \text{BOC}=180^\circ$
$\angle \text{BOC}=180^\circ-60^\circ$
$\angle \text{BOC}=120^\circ$

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