Question 109

$$ABCD$$ is a cyclic quadrilateral such that $$AB$$ is a diameter of the circle circumscribing it and angle $$ADC = 125^\circ$$. Then angle $$BAC$$ is equal to:

Solution

Sum of opposite angles of a quadrilateral is 180°.

So,$$\angle ADC+\angle ABC=180°.$$

or,$$\angle ABC=55°.$$

Again for semicircle,

$$\angle ACB= 90°.$$

$$\angle BAC =$$

$$180-\angle ACB-\angle ABC$$

or,$$\angle BAC=180-90-55=35°$$

D is correct choice.

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