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

Quantitative Aptitude (CGL) Geometry

ABCD is a cyclic quadrilateral such that AB is the diameter of the circle circumscribing it and $\angle$ ADC = $129^{\circ }$ . Then, $\angle$ BAC is equal to:

This questions was previously asked in

SSC CGL 13 June 2019 Shift-3

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$61^{\circ }$

$49^{\circ }$

$39^{\circ }$

$51^{\circ }$

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Explanation:

As per the given question,
It is given that $\angle$ ADC = $129^{\circ }$ .

As per the diagram, ABCD is a cyclic quadrilateral. So $\angle$ ADC + $\angle$ ABC = 180
Now, $\angle$ ABC = 180 - $129^{\circ }$ = $51^{\circ }$
Now, In $\bigtriangleup$ ACB
$\angle$ ACB = $90^{\circ }$ (angle at the circumference with in half circle)
So, $\angle$ BAC = $90^{\circ }$ - $51^{\circ }$ = $39^{\circ }$