$O$ is the circumcenter of non-right $\triangle ABC$. $\frac{|AB \cdot CO|}{|AC \cdot BO|} = \frac{|AB \cdot BO|}{|AC \cdot CO|} = 3$. Find $\tan A$ Problem: $O$ is the circumcenter of $\triangle ABC$, which is not a right triangle.  $$\frac{| AB \cdot CO |}{|AC \cdot BO|} = \frac{|AB \cdot BO|}{|AC \cdot CO|} = 3$$. Find $\tan A$. Here $AB \cdot CO$ represents the dot product of vector $\overrightarrow{AB}$ and $\overrightarrow{CO}$
My thoughts: WLOG let the radius of the excircle be $1$. Let $AB = c$, $BC = a$, $AC 
= b$. Then $AB \cdot CO = c \cdot 1 \cdot \cos( 90^{\circ} + \angle B- \angle A ) = -c \cdot \sin( \angle B- \angle A)$
$AC \cdot BO = -b \cdot \sin( \angle C- \angle A)$
$AB \cdot BO = c \sin C$
$AC \cdot CO = b \sin B$
But none of these look very trivial to solve..
 A: Let us record as in the OP the values for the scalar products, then start the computation using the formulas $a=2R\sin A$ (and the similar ones for $b,c$) to express the appearing sines, and $b^2+c^2-a^2=2bc\cos A$ (and the similar ones) to express the appearing cosines.
$$
\begin{aligned}
|AB\cdot CO|
&= AB\cdot CO\cdot\cos(\widehat{AB,CO})=cR\; \cos\left(\frac \pi2-A+B\right)
=cR\sin(A-B)\ ,
\\
|AC\cdot BO|
&= AC\cdot BO\cdot\cos(\widehat{AC,BO})=bR\; \cos\left(\frac \pi2-A+C\right)
=bR\sin(A-C)\ ,
\\[4mm]
|AB\cdot BO|
&= AB\cdot BO\cdot\cos(\widehat{AB,BO})=cR\; \cos\left(\frac \pi2-C\right)
=cR\sin C=2R^2\sin^2 C\ ,
\\
|AC\cdot CO|
&= AC\cdot CO\cdot\cos(\widehat{AC,CO})=bR\; \cos\left(\frac \pi2-B\right)
=bR\sin B=2R^2\sin^2 B\ .
\end{aligned}
$$
From the last two equalities, and the given second proportion, we get
$$
3=\frac{|AB\cdot BO|}{|AC\cdot CO|}
=\frac{\sin^2 C}{\sin^2 B}=\frac {c^2}{b^2}\ .
$$
So $c=b\sqrt 3$.
We may want to norm $b=2$, so then $c=2\sqrt 3$.
Let us use the first condition...
$$
\begin{aligned}
\pm 3 
&=
\frac{|AB\cdot CO|}{|AC\cdot BO|}
\\
&=\frac{cR\sin(A-B)}{bR\sin(A-C)}
=\frac cb\cdot\frac{\sin A\cos B-\cos A\sin B}{\sin A\cos C-\cos A\sin C}
\\[2mm]
&=
\frac cb\cdot
\frac
{\displaystyle a\cdot\frac {a^2+c^2-b^2}{2ac} - \frac {b^2+c^2-a^2}{2bc} \cdot b}
{\displaystyle a\cdot\frac {a^2+b^2-c^2}{2ab} - \frac {b^2+c^2-a^2}{2bc} \cdot c}
\\[2mm]
&=
\frac
{(a^2+c^2-b^2) - (b^2+c^2-a^2)}
{(a^2+b^2-c^2) - (b^2+c^2-a^2)}
\\[2mm]
&=
\frac
{a^2-b^2}
{a^2-c^2}\ .
\end{aligned}
$$
Recall the norming making $b^2=4$, $c^2=3b^2=12$.
The two chances for $a^2$ are now $10$ and $16=4+12$, where the above becomes
$-3=\frac{10-4}{10-12}$
and
$+3=\frac{16-4}{16-12}$. The second case gives rise to a triangle with $A=90^\circ$, which was excluded in the title. So we have to work with the triangle with sides:
$$
\color{blue}{
\boxed{\qquad a=\sqrt{10}\ ,\ b=2\ ,\ c=2\sqrt 3=\sqrt{12}\ .\qquad}
}
$$
Let us compute $\tan A$ by explicitly computing $\cos A$, $S$, $R$, $\sin A$ in this order:
$$
\begin{aligned}
\cos A&=\frac{b^2+c^2-a^2}{2bc}=\frac{4+12-10}{2\cdot 2\cdot 2\sqrt 3}=\frac 6{8\sqrt 3}=\color{blue}{\frac{\sqrt 3}4}\ ,
\\[2mm]
S^2&=s(s-a)(s-b)(s-c)\qquad\text{(Heron)}
\\
&=\frac 1{16}(a+b+c)(b+c-a)(a+c-b)(a+b-c)
=\frac 1{16}\Big((b+c)^2-a^2\Big)\Big(a^2-(b-c)^2\Big)
\\
&=\frac 1{16}\Big(2bc + (b^2+c^2-a^2)\Big)\Big(2bc - (b^2+c^2-a^2)\Big)
=\frac 1{16}\Big(4b^2c^2 -(b^2+c^2-a^2)^2\Big)
\\
&=\frac 1{16}\Big(4\cdot 4\cdot 12 -6^2\Big)
=\frac{39}4\ ,
\\
S&=\frac {\sqrt{39}}2\ ,
\\[2mm]
R&=\frac{abc}{4S}=2\sqrt \frac{10}{13}\ ,
\\[2mm]
\sin A&=\frac a{2R}=\frac {\sqrt {10}}{4\sqrt {10/13}} =\color{blue}{\frac{\sqrt{13}}4}\ ,
\\
\tan A&=\frac{\sin A}{\cos A}=\frac{\sqrt{13}/4}{\sqrt 3/4}
=\color{blue}{\boxed{\ \sqrt{\frac{13}3}}\ }\ .
\end{aligned}
$$
$\square$
