Angle in an isosceles triangle formed by circle radii

In $\triangle AOB$, $OA = OB = 6\text{ cm}$ (radii of circle). Thus $\triangle AOB$ is isosceles with $\angle OAB = \angle ABO$. Sum of angles in $\triangle AOB = 180^\circ$. $\angle AOB + 2 \angle ABO = 180^\circ \implies 114^\circ + 2 \angle ABO = 180^\circ \implies \angle ABO = \frac{180^\circ - 114^\circ}{2} = 33^\circ$.