Question

In the given figure, O is the centre of a circle.If $\angle$DAC = 54${}^o$ and $\angle$ACB = 63${}^o$ then $\angle$BAC =

• A. 72${}^o$
• B. 54${}^o$
• C. 27${}^o$
• D. 90${}^o$

Answer 1

Answer: (C)

27${}^o$

Given- ABCD is a quadrilateral inscribed in a circle with centre O which is on the line AC.
$\angle ACB={63}^o$ and $\angle DAC={54}^o$.
To find out- $\angle BAC=?$
Solution- The quadrilateral has been inscribed in the circle.
$\therefore$ The points A, B, C & D are on the circumference of the circle. Again the centre O is on the line AC.
$\therefore$ AC is the diameter of the circle and the angle at the centre O is a straight one, i.e. $\angle AOC={180}^o$. Now $\angle ADC\&\angle ABC$ are angles in a semicircle. $\therefore \angle ADC={90}^o=\angle ABC$.
So in $\Delta ABC\angle BAC={180}^o-(\angle ABC+\angle ACB)={180}^o-({90}^o+{63}^o)={27}^o$.
Ans-Option C.