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 $$\angle BOC$$


ANSWER

$$55^o$$


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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