Question 94

In quadrilateral $$ABCD$$, the bisectors of $$\angle A$$ and $$\angle B$$ meet at $$O$$ and $$\angle AOB = 64^\circ. \angle C + \angle D$$ is equal to:

Solution

In $$\triangle$$ AOB,
$$\angle OAB + \angle OBA + \angle O = 180$$
$$\angle OAB + \angle OBA = 180 - 64 = 116\degree$$
$$\angle$$ OAB and $$\angle OBA$$ is the bisector of $$\angle$$ A and $$\angle$$ B.
So,
$$\angle$$ A + $$\angle$$ B = 2 $$\times$$ 116 = 232$$\degree$$
$$\angle$$ A + $$\angle$$ B + $$\angle$$ C + $$\angle$$ D = 360
$$\angle$$ C + $$\angle$$ D = 360 - 232 = 128$$\degree$$

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SSC CGL Tier-2 11th September 2019 Maths

In quadrilateral $$ABCD$$, the bisectors of $$\angle A$$ and $$\angle B$$ meet at $$O$$ and $$\angle AOB = 64^\circ. \angle C + \angle D$$ is equal to:

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