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 =130^o . Then angle BAC is

equal to:

This questions was previously asked in

SSC CPO 14th-March-2019-Shift-2

150^o

40^o

50^o

60^o

Let’s break it down:

- AB is a diameter, so angle at the circle on AB (other than AB itself) is 90°, thanks to the angle in a semicircle rule.

- ABCD is cyclic, so opposite angles add to 180°.

- Given: ?ADC = 130°.

- By cyclic property, ?ABC + ?ADC = 180°, so ?ABC = 50°.

- Now, in triangle ABC, since AB is diameter, ?ACB = 90°.

- So, in triangle ABC: Angles are ?ABC = 50°, ?ACB = 90°, means ?BAC = 40°.

- So from the options:

1. 150° — way too big.

2. 40° — matches our calculation.

3. 50° — this is ?ABC, not ?BAC.

4. 60° — not aligning with the properties here.

- The answer is option 2: 40°.

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