Mathematics

In the triangle ABC, the bisectors of exterior angles B and C meet at O. Given that $\angle BAC = 70^{\circ}$ and $\angle ACB C = 50^{\circ}$, find $\angle BOC$

$55^o$

SOLUTION
In $\triangle ABC$,
$\angle ABC + \angle ACB + \angle BAC = 180$
$\angle ABC + 70 + 50 = 180$
$\angle ABC = 60$

$\angle OCB = \frac{1}{2} (180 - \angle ACB)$
$\angle OCB = \frac{1}{2}(180 - 50)$
$\angle OCB = 65^{\circ}$

$\angle OBC = \frac{1}{2} (180 - \angle ABC)$
$\angle OBC = \frac{1}{2}(180 - 60)$
$\angle OBC = 60^{\circ}$

In $\triangle OBC$,
$\angle OCB + \angle OBC + \angle BOC = 180$
$65 + 60 + \angle BOC = 180$
$\angle BOC = 180 - 65 - 60$
$\angle BOC = 55^{\circ}$

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Single Correct Medium Published on 09th 09, 2020
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