geometry

Given that point O is the center of the circle and OA=5, OC=8, and \(\angle ACO=30^\circ\), find \(\angle BOA\) in degrees.

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{\sin\left(90^\circ+\dfrac{x}{2}\right) }{8}} &=& \mathbf{\dfrac{\sin(30^\circ)}{5}} \quad | \quad \sin\left(90^\circ+\dfrac{x}{2}\right) = \cos\left( \dfrac{x}{2} \right),\ \sin(30^\circ) = \dfrac{1}{2} \\\\ \dfrac{\cos\left( \dfrac{x}{2} \right) }{8} &=& \dfrac{\dfrac{1}{2}}{5} \\\\ \dfrac{\cos\left( \dfrac{x}{2} \right) }{8} &=& \dfrac{1}{10} \\\\ \cos\left( \dfrac{x}{2} \right) &=& \dfrac{8}{10} \\\\ \cos\left( \dfrac{x}{2} \right) &=& \dfrac{4}{5} \\\\ \dfrac{x}{2} &=& \arccos\left( \dfrac{4}{5} \right) \\\\ \dfrac{x}{2} &=& 36.8698976458^\circ \\ x &=& 2* 36.8698976458^\circ \\ \mathbf{x} &=& \mathbf{73.7397952917^\circ} \\ \hline \end{array}\)

\(\mathbf{\angle BOA \approx 73.74^\circ}\)