Plane Geometry problem I came across this problem in a mathematics-related facebook group. Could anyone advise on the solution to it(i.e. hints only)? Thank you.

 A: If the circle is centered at (0,0) then the top right vertex of the triangle is at
$$\left[\frac{2}{\sqrt{3}},1\right]$$
and the circle intersects the triangle at
$$\left[\frac{\sqrt 3} 2,\frac1 2 \right]$$
Substitute to get $$\cot \phi = 3 \sqrt{3}$$
A: My solution:
Take a circle centered at (0,0) and radius r.
Now build a triangle as shown in the figure and you will get its vertices as $(0,r),(-\frac{2r}{\sqrt3},-r),(\frac{2r}{\sqrt3},-r)$
Now solve one of the sides of triangle(except the one which is tangent) with the circle.
You will get its coordinates as $(-\frac{r\sqrt3}2,-\frac{r}2)$
Now find the slopes of the lines(which contain the angle $\phi$), say $m_1$ and $m_2$
Now find $\tan \phi = |\frac{m_1 - m_2}{1 + m_1m_2}|$
From here find $\cot \phi = \frac1{\tan \phi}$
You will get that as $3\sqrt3$
A: 
Since triangle ABC is equilateral, we know that angle BAC is 60º.  Therefore, angle PAQ is also 60º; we have a circle theorem that tells us that angle POQ is thus twice that measure or 120º.  The diameter AD bisects that angle, so angle POD is 60º.
Diameter AD is also an altitude of equilateral triangle ABC:  thus, if  $ \ s \ $ is the length of one side of that triangle, the diameter of the circle is $ \ \frac{\sqrt{3}}{2}s \ $ . The radius OD therefore has length $ \ \frac{\sqrt{3}}{4}s \ $ .  Since the altitude AD bisects side BC, then the segment BD has length $ \ \frac{1}{2}s \ $ .  Hence, $ \ \tan(\angle BOD ) \ = \ \frac{2}{\sqrt{3}} \  $ .
Now, the measure of angle POB, designated as $ \ \phi \ , $ is the difference given by the measure of angle POD (call it $ \ \alpha \ $ ) minus the measure of angle BOD (call that $ \ \beta \ $ ) .  We can then use the formula for the tangent of the difference of two angles, with $ \ \tan(\angle POD) \ = \ \tan(\alpha) \ = \ \tan \ 60º \ = \ \sqrt{3} \ $ , to obtain
$$ \tan \phi \ = \ \tan( \alpha - \beta ) \ = \ \frac{\tan \alpha \ - \ \tan \beta}{1 \ + \ \tan \alpha \cdot \tan \beta} \ = \ \frac{\sqrt{3} \ - \ \frac{2}{\sqrt{3}}}{1 \ + \ \sqrt{3} \cdot \frac{2}{\sqrt{3}}} \ = \ \frac{ \frac{3 \ - \ 2}{\sqrt{3}}}{1 \ + \ 2 } \ = \ \frac{1}{3 \ \sqrt{3}} \ .  $$
We thereby conclude that  $ \ \cot \phi \ = \ 3 \ \sqrt{3} \ . $
