Angle in a triangle within a circle. A and B are two points on the circumference of a circle with centre O. C is a point on OB such that AC $\perp OB$. AC = 12 cm. BC = 5 cm. Calculate the size of $\angle AOB$, marked $\theta$ on the diagram.

The answer given in the textbook is $45.2 ^\circ$ (1dp)
Note: This is not a homework question. I'm just doing maths for my own interest.

let radius = $r$, then in triangle ACO, using Pythagoras:
$$\begin{align}
AO^2 &= AC^2+CO^2\\
\\
r^2 &= 12^2+(r-5)^2\\
\\
144+25-10r &= 0\\
\\
r &= 16.9\\
\end{align}$$
In triangle ACO,
$$\begin{align}
\sin\theta &= \dfrac{12}{r}\\
\\
\sin\theta &= \dfrac{12}{16.9}\\
\\
\theta &= 45.2^{\circ} (1dp)\\
\end{align}$$
