Question

In the given figure, find x+y

• A. ${116}^{\circ }$
• B. ${102}^{\circ }$
• C. ${64}^{\circ }$
• D. ${76}^{\circ }$

Answer 1

Answer: (B)

${102}^{\circ }$

For cyclic quadrilateral ABCD
$y+{52}^{\circ }={116}^{\circ }\Rightarrow y={116}^{\circ }-{52}^{\circ }={64}^{\circ }$
(Ext, angle property of cyclic quadrilateral)
OB=OA (radii of same circle)
$\Rightarrow \angle OAB=\angle OBA={52}^{\circ }$
$\therefore \angle BOA={180}^{\circ }-(\angle OAB+\angle OBA)$
$={180}^{\circ }-({52}^{\circ }+{52}^{\circ })$
$={180}^{\circ }-{104}^{\circ }={76}^{\circ }$(Angle sum property of a $△$)
$\therefore x=\angle BCA=\frac{1}{2}\times \angle BOA=\frac{1}{2}\times {76}^{\circ }={38}^{\circ }$
(Angle at the centre is twice the angle at any remaining parts of circumference)
$\therefore x+y={64}^{\circ }+{38}^{\circ }={102}^{\circ }$