Question 82
- ΔABC is isosceles having AB = AC and $$ \angle A $$ = $$40^\circ$$. Bisectors PO and OQ of the exterior angles $$ \angle ABD $$and $$ \angle A CE $$ formed by producing BC on both sides, meet at O. Then the value of $$ \angle BOC $$ is
Solution
AB=AC
Therefore, $$ \angle ABC = \angle ACB = \frac{140^\circ}{2} = 70^\circ $$
Therefore, $$ \angle ABD = \angle ACE = 180^\circ-70^\circ=110^\circ $$
Therefore, $$ \angle PBD = 55^\circ = \angle CBO $$
$$ \angle QCE = \angle BCO = 55^\circ $$
Therefore, $$ \angle BOC = 180^\circ - (2 \times 55^\circ ) = 70^\circ $$
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