$\triangle \mathrm{ABC}$ is incribed in a circle with center $\mathrm{O}$, such that $\angle \mathrm{ACB}=115^{\circ}$. O is joined to $\mathrm{A}$. What is the measure of $\angle \mathrm{OAB}$ ?

25°

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of a circle.

= External \(\angle\)AOB = 2\(\angle\)ACB

= External \(\angle\)AOB = \({115}^\circ\) x 2 = \({230}^\circ\)

= Internal \(\angle\)AOB = \({360}^\circ\) - \({230}^\circ\) = \({130}^\circ\)

In \(\Delta \)ABO, AO = BO = Radii of circle.

= \(\angle\)BAO = \(\angle\)ABO

In \(\Delta \)ABO,

= \(\angle\)BAO + \(\angle\)ABO + \(\angle\)AOB = \({180}^\circ\)

= \(\angle\)BAO = \(\frac{1}{2}\) x (\({180}^\circ\) - \({130}^\circ\)) = \({25}^\circ\)

Therefore, \(\angle\)BAO = \({25}^\circ\)