Question

In the given figure, AB is the chord to the circle having centre at O. If AC = CB and $\angle AOC={50}^{\circ }$, then find $\angle CAO$.

Answer 1

In the given figure AB is the chord and AC = CB.
Since OC is bisecting the chord AB.
In Δ OCA and Δ OCB, we have:
AC = BC
(Given)
OC = OC
(Common)
OA = OB
(Radii of a circle)
∴ Δ OCA ≅ Δ OCB (By SSS congruency rule)
⇒ ∠ OAC = ∠ OBC (CPCT)
⇒ ∠ AOC = ∠ BOC (CPCT)
Hence, $\angle BOC={50}^{\circ }$
$\angle AOB={100}^{\circ }$
Now in $\Delta OAB$
$\angle AOB={100}^{\circ }$
∠ OAB = ∠ OBA
So, $2\angle OAB={180}^{\circ }-\left({100}^{\circ }\right)={80}^{\circ }$
$\angle OAB={40}^{\circ }$
⇒ $\angle CAO={40}^{\circ }$