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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Subjective Medium Published on 09th 09, 2020
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