The distance from the incenter of the triangle to the vertex B

Question

For reference: The angle $B$ of a triangle $ABC$ measures $16^o$,
its perimeter is $56$ and $AB \cdot BC=600$.
Calculate the distance from the incenter of the triangle to the vertex $B$.(answer:$15\sqrt2$

My progress:

I know the formula for the distance from the incenter to the vertex:
$AB = c, BC = a$ and $AC = b$

$IB=\sqrt{\frac{(p-b).ac}{p}}=\sqrt{\frac{(56-b)600}{56}}=\sqrt{\frac{(56-b).75}{7}}$

but I don't know where to put the angles as they are not noticeable..and whether it is possible to solve by geometry?

Another idea:

Answer 1

$$BI=\frac{r}{\sin8^{\circ}}=\frac{\frac{2S}{a+b+c}}{\sin8^{\circ}}=\frac{ac\sin16^{\circ}}{(a+b+c)\sin8^{\circ}}=\frac{600\cdot2\cos8^{\circ}}{56}=...$$

So we did not get your answer.