Here we have $\triangle ABC$ with circumcenter $O$ , Incenter $I$ and excenter opposite to vertex $A$ as $E$.As the title suggests, I have to find $EO$.
Now I know $OI=R\sqrt{1+8\Pi_{cyc}\sin (A/2)}$ and $IE=4R\sin (A/2)$. So I want to find third side and I know two sides , therefore I'm thinking of using cosine rule in $\triangle EOI$ but for that I require to find one angle of the $\triangle EOI$ for using cosine rule and I'm not able to figure out it.
Can You provide me some hints?
NOTE: I've already checked the existing answer for this question but I want to do it y cosine rule.
We know that $IA=r\csc(\frac{A}{2})=4R\sin \frac{B}{2}\sin \frac{C}{2}$, $~~OI=\sqrt{R^2-2Rr}=R\sqrt{1-8\Pi_{cyc}\sin (A/2)}$, and $OA=R$
$$\implies \cos \theta=\frac{OI^2+IA^2-R^2}{2(OI)(IA)}$$ $$OE^2=OI^2+IE^2+2(OI)(IE)\cos \theta=OI^2+IE^2+\frac{(IE)(OI^2+IA^2-R^2)}{IA}$$