Austin's angle

If you draw it correctly,

you can notice that $BO$ cuts $∠ABC$ right in the center.

So now, there will be new two angles that measure the same,

$∠ABO = ∠CBO = 16^∘$ .

Since $BO$ and $CO$ are $R$ ,

we can respectively say that $BO = CO$ ,

so we get new isosceles triangle, $BCO$ .

We now that base angles of isosceles measure the same.

That means $∠CBO = ∠BCO = 16^∘$ .

Sum of all angles of triangle will always $180^∘$ ,

then we put some of our math:

$∠BCO + ∠CBO + ∠BOC = 180^∘$

$16^∘ + 16^∘ + ∠BOC = 180^∘$

$32^∘ + ∠BOC = 180^∘$

$∠BOC = 180^∘ - 32^∘$

$∠BOC = 148^∘$

the mesure of the angle <BOC is the double of the angle <BAC . Since the some of mesures of angle of triangle is 180° and ABC isosceles and <ABC is 32° we can deduce that mesure of <BAC is 74° then mesure of <BOC is 148°

One of the base angles of the triangle $\angle{BAC}$ is the inscribe angle of $\circ{O}$ . Thus, $\angle{BOC}$ = 2 $\angle{BAC}$ . $\angle{BOC}$ = $\text{148}$ .