Question

In the given figures, O is the centre of the given circles. Match the values of $x$.

• A. ${40}^{\circ }$
• B. ${60}^{\circ }$
• C. ${200}^{\circ }$

Answer 1

In the first image, we see that AB=CD.
We know that equal chords subtend equal angles at the centre.
$⟹\angle AOB=\angle COD={40}^{\circ }$

In the second image, AB=CD and $\angle AOB={60}^{\circ }.$
Since equal chords subtend equal angles at the centre, we have
$\angle AOB=\angle COD={60}^{\circ }.$
Note that $△s$ AOB and COD are isosceles.
Thus, using angle sum property in $△COD,$ we have
$\angle COD+\angle OCD+\angle ODC={180}^{\circ }.$
$⟹{60}^{\circ }+x+x={180}^{\circ }$
$⟹x={60}^{\circ }$

In the third image, since $△OBC$ is isosceles, we have $\angle OBC=\angle OCB={40}^{\circ }.$
Now, using angle sum property in $△OCB,$ we have
${40}^{\circ }+{40}^{\circ }+\angle BOC={180}^{\circ }⟹\angle BOC={100}^{\circ }.$
Now, since AB and CB are equal chords, and equal chords subtend equal angles at the centre,
$\angle AOB=\angle BOC={100}^{\circ }.$
Thus, $x=\angle AOB+\angle BOC={200}^{\circ }.$