Mathematics

# In a $\triangle ABC$, the sides AB and AC have been produced to D and E. Bisectors of $\angle CBD$ and $\angle BCE$ meet at O. If $\angle A={ 64 }^{ 0 }$, then $\angle BOC$ is

${ 58 }^{ 0 }$

##### SOLUTION
Given: OB and OC bisect $ext. \angle B$ and $ext. \angle C$, $\angle A = 64^{\circ}$

Now, In $\triangle OBC$,
Sum of angles = 180
$\angle OBC + \angle OCB + \angle BOC = 180$
$\frac{1}{2} (ext. \angle B + ext. \angle C) + \angle BOC = 180$ (OB and OC bisect exterior angles)
$\frac{1}{2} (180 - \angle ABC + 180 - \angle ACB) + \angle BOC = 180$
$\frac{1}{2} (360 - (\angle ABC + \angle ACB)) + \angle BOC = 180$
$\frac{1}{2} (360 - (180 - \angle A)) + \angle BOC = 180$ (Angle sum property)
$\frac{1}{2} (180 + \angle A) + \angle BOC = 180$
$\angle BOC = 180 - 90 -\frac{1}{2} (64)$
$\angle BOC = 58^{\circ}$

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Single Correct Medium Published on 09th 09, 2020
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