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 


ANSWER

$${ 58 }^{ 0 }$$


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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