sinA/a

$\dfrac{5}{\sin 30\degree}=\dfrac{8}{\sin \angle {OBC}}=\dfrac{8}{\sin \angle {OBA}}$

$\implies \angle {OBA}=\sin^{-1} (\frac{4}{5})\approx 53.1301\degree$

$\implies \angle {AOB}=180\degree-2\times \angle {OBA}\approx \boxed {73.7398\degree}$ .

Draw the median $OM$ of $\triangle AOB$ where $M$ is the midpoint of $AB$ .

Since $\triangle AOB$ is isosceles, $\angle OMB$ is a right angle, and solving right triangle $\triangle OMC$ with $OC = 8$ and $\angle ACO = 30°$ gives $OM = 4$ , and solving right triangle $\triangle MOA$ with $OM = 4$ and $OA = 5$ gives $\angle MOA = \cos^{-1} \frac{4}{5}$ .

Therefore, $\angle AOB = 2 \angle MOA = 2 \cos^{-1} \frac{4}{5} \approx \boxed{73.7398°}$ .

We can apply the cosine law on both the triangles COB and COA to find the missing side then subtract AC-CB to get AB and then use that to find the angle as we have the 3 sides in the triangle ABO with the cos law

$\angle OAB^{\circ} = \angle OBA^{\circ}$

In $\triangle AOC$ : $\dfrac{5}{\sin{(30^{\circ})}} = \dfrac{8}{\sin{(\angle OAC^{\circ})}} \implies \angle OAC^{\circ} \approx 53.1$

$\begin{aligned} \angle BOA^{\circ} &= 180^{\circ} - \angle OAB^{\circ} - \angle OAB^{\circ} \\ \angle BOA^{\circ} &= 180^{\circ} - 2 (53.1^{\circ}) \\ \angle BOA^{\circ} &= \boxed{73.73^{\circ}} \end{aligned}$

Using sine law on triangle OBC,we get $sin(ABO)=\frac{4}{5}$

And the answer follows immediately using the fact that the sum of the angles of a triangle sum up to 180 degree and OA=OB

Using the sine rule,
$\frac{sin \angle ACO}{OA}=\frac{sin \angle OAC}{OC}.$
So, $\frac{0.5}{5}$ = $\frac{sin \angle OAC}{8}$ , i.e. $sin\angle OAC$ =0.8.
Therefore, $\angle OAC = sin^{-1}0.8=53.13010...$
Since the interior angles of a triangle add up to $180^\circ$ and OA & OB are the radii of the same circle i.e. triangle OAB is isosceles,
$\angle AOB=180^\circ - 2\times\angle OAC=73.739795...$ So the answer is 73.739795...