Mathematics

In a triangle ABC, the angle bisectors of angle ABC and ACB meet at point O. If $$\angle$$BAC $$=32^o$$, find $$\angle$$ BOC.


SOLUTION
Let the $$\angle OBC = x$$ and $$\angle OCB = y$$
$$\angle BOC = 180-(x+y)$$
Exterior $$\angle BOC = 360 - (180-(x+y)) = 180 + (x+y)$$
For the quadilateral $$ABOC$$, we have 
$$32+180+(x+y)+x+y = 360$$
$$2(x+y) = 148$$
$$(x+y) = 74$$
$$\angle BOC = 180-74 = 106^o$$
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