Question 65
In a quadrilateral $$ABCD$$, the bisectors of $$\angle C$$ and $$\angle D$$ meet at $$E$$. If $$\angle CED = 56^\circ$$ and $$\angle A = 49^\circ$$, then the measure of $$\angle B$$ is:
Solution
In $$\triangle CED$$,
$$\angle EDC + \angle ECD + \angle CED = 180$$
$$\angle EDC + \angle ECD = 180 - 56 = 124$$
$$\angle C +Â \angle D = 2(\angle EDC + \angle ECD)$$
($$\because$$ angle bisector.)
$$\angle C + \angle D = 2 \times 124 = 248$$
In quadrilateral $$ABCD$$,
$$\angle A +Â \angle B +Â \angle C +Â \angle D = 360$$
$$\angle 49 + \angle B + 248 = 360$$
$$\angle B = 360 - 297 = 63\degree$$
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