Question

$A,B,C$ are point in the circle with centre $O$. If $OCA=x$ then find $\angle OAC$. Prove that $\angle OCA$ +$\angle ABC={90}^o$.

Answer 1

$\begin{array}{c}\therefore \angle AOC=2\angle ABC\left(byproperities\right)\\ and,\angle OCA=\angle OAC=x\left[OA=OC\right]\\ Nowin\Delta AOC\\ \angle AOC+\angle OAC+\angle OCA={180}^{\circ }\\ \Rightarrow 2\angle ABC+x+x={180}^{\circ }\\ \Rightarrow 2\left(\angle ABC+x\right)={180}^{\circ }\\ \Rightarrow \angle ABC+\angle OAC={90}^{\circ }.\\ poved.\end{array}$