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:

$$\begin{align}
AO^2 &amp;= AC^2+CO^2\\
\\
r^2 &amp;= 12^2+(r-5)^2\\
\\
144+25-10r &amp;= 0\\
\\
r &amp;= 16.9\\
\end{align}$$

In triangle ACO,
$$\begin{align}
\sin\theta &amp;= \dfrac{12}{r}\\
\\
\sin\theta &amp;= \dfrac{12}{16.9}\\
\\
\theta &amp;= 45.2^{\circ} (1dp)\\
\end{align}$$
Draw line AB, and observe that ABC is a (5,12,13) right-angled triangle. Since AB is a chord, the bisector of angle AOB also bisects AB, call this point T. Now triangle OTB is similar to triangle ACB. Therefore angle AOB is:

2 * arctan(5/12)

Which a handy calculator says is 45.2 degrees.

(I'm surprised someone marked this down: I did get the answer, which one other answer didn't, and I think it's neater than the other answer, since you need almost no calculation.)