Question

In the above figure, $\overline{AB}$ and $\overline{AC}$ are equal chords and $O$ is the centre. If $\angle BOC={100}^o$, then find $\angle ACO$.

• A. ${10}^o$
• B. ${20}^o$
• C. ${30}^o$
• D. ${25}^o$

Answer 1

The correct option is D ${25}^o$

Consider the diagram shown in the question.

Given:

$AB=AC$

$\angle BOC=100{}^{\circ }$

$\Rightarrow \angle BAC=50{}^{\circ }$ (angle at the circle is half of angle at the centre by same chord $\left(BC\right)$)

Consider the quadrilateral $OBAC$.

$\angle BOC$ (of the quadrilateral) $=360{}^{\circ }-100{}^{\circ }=260{}^{\circ }$

Consider triangle $OBC$.

$\angle OBC=\angle OCB$ ($OB$ and $OC$ are radius)

Similarly, in $\Delta ABC$,

$\angle ABC=\angle ACB$ (Since, $AB=AC$)

$\angle ABO+\angle OBC=\angle ACO+\angle OCB$

$\Rightarrow \angle ABO=\angle ACO$

We know that interior angles of a quadrilateral are $360{}^{\circ }$. Therefore,

$50{}^{\circ }+260{}^{\circ }+\angle ABO+\angle ACO=360{}^{\circ }$

$2\angle ACO=50{}^{\circ }$

$\angle ACO=25{}^{\circ }$

Hence, this is the required result.