ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and angle ...........

Quantitative Aptitude (CPO) Geometry

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and angle ADC = 148^o. Then angle BAC is

equal to:

This questions was previously asked in

SSC CPO 15th-March-2019-Shift-2

View All SSCCPO Previous Year Papers

60^o

58^o

40^o

150^o

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Explanation:

Let’s break this down step by step:

- AB is the diameter of the circle. That means angle BAC is an angle in a semicircle, so it must be a right angle (90°).

- In any cyclic quadrilateral, the sum of the opposite angles is 180°. So:

?BAD + ?ADC = 180°

- They gave you ?ADC = 148°. So:

?BAD = 180° – 148° = 32°

- But they’re asking for angle BAC. Remember, BAC is at the end of the diameter, so it’s definitely 90°, no matter what the other angles do.

- Let’s check the options:

1. 60°

2. 58°

3. 40°

4. 150°

- None of the answers is 90°, which is odd, but it’s clear BAC cannot ever be 58°. That’s not how angles in a semicircle work.

So, the correct value for angle BAC is 90°, but if you wanted to pick from the options, none would be right. Out of the given choices, none matches. But for this particular problem, if "BAC is the angle in the semicircle," it should be 90°. Just remember: angle in a semicircle? Always 90°.