Using cosine rule to find distance between circumcenter and excenter

Question

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.

Wish the downvoter had left a comment – Lalit Tolani Jun 11 '21 at 12:49 — Lalit Tolani, Jun 11 '21 at 12:49

Asher2211 · Accepted Answer · 2021-06-12T03:31:15.253

0

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 $OA=R$
$$\implies \cos \theta=\frac{OI^2+IA^2-R^2}{2(OI)(IA)}$$ $$OE^2=OI^2+IE^2+2(OI)(IE)\cos \theta=OI^2+IE^2+\frac{(IE)(OI^2+IA^2-R^2)}{IA}$$

edited Jun 12 '21 at 03:31

answered Jun 11 '21 at 10:00

Asher2211

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I'm getting answer as $OE^2=R^2(1+24\sin (A/2)\cos (B/2)\cos (C/2))$ which is incorrect, Can you please check once – Lalit Tolani Jun 11 '21 at 13:59
looking for a reply – Lalit Tolani Jun 12 '21 at 03:21
I have edited my answer $OI=R\sqrt{1-8\Pi_{cyc}\sin (A/2)}$ and not $R\sqrt{1+8\Pi_{cyc}\sin (A/2)}$. Can you solve from here? – Asher2211 Jun 12 '21 at 03:32

Using cosine rule to find distance between circumcenter and excenter

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