Let $a$, $b$ and $c$ be positive real numbers. Consider a triangle ABC whose vertices are the three points $\mathrm{A}(a, 0)$, $\mathrm{B}(3, b)$ and $\mathrm{C}(0, c)$ on a plane with the coordinate system. Assume that the circumscribed circle of the triangle ABC passes through the origin $\mathrm{O}(0,0)$ and that $\angle\mathrm{BAC} = 60^\circ$. (1) Since $\angle\mathrm{AOB} = \mathbf{AB}^\circ$, we obtain $b = \sqrt{\mathbf{C}}$. (2) The equation of the circumscribed circle is $$\left(x - \frac{a}{\mathbf{D}}\right)^2 + \left(y - \frac{c}{\mathbf{E}}\right)^2 = \frac{a^2 + c^2}{\mathbf{F}},$$ and $c$ can be expressed in terms of $a$ as $c = \sqrt{\mathbf{G}}\,(\mathbf{H} - a)$. (3) Let D denote the point of intersection of the segment OB and the segment AC. Set $\alpha = \angle\mathrm{OAC}$ and $\beta = \angle\mathrm{ADB}$. When $a = 2\sqrt{3}$, it follows that $$\tan\alpha = \mathbf{I} - \sqrt{\mathbf{J}}, \quad \tan\beta = \mathbf{K}.$$
Let $a$, $b$ and $c$ be positive real numbers. Consider a triangle ABC whose vertices are the three points $\mathrm{A}(a, 0)$, $\mathrm{B}(3, b)$ and $\mathrm{C}(0, c)$ on a plane with the coordinate system. Assume that the circumscribed circle of the triangle ABC passes through the origin $\mathrm{O}(0,0)$ and that $\angle\mathrm{BAC} = 60^\circ$.
(1) Since $\angle\mathrm{AOB} = \mathbf{AB}^\circ$, we obtain $b = \sqrt{\mathbf{C}}$.
(2) The equation of the circumscribed circle is
$$\left(x - \frac{a}{\mathbf{D}}\right)^2 + \left(y - \frac{c}{\mathbf{E}}\right)^2 = \frac{a^2 + c^2}{\mathbf{F}},$$
and $c$ can be expressed in terms of $a$ as $c = \sqrt{\mathbf{G}}\,(\mathbf{H} - a)$.
(3) Let D denote the point of intersection of the segment OB and the segment AC.
Set $\alpha = \angle\mathrm{OAC}$ and $\beta = \angle\mathrm{ADB}$. When $a = 2\sqrt{3}$, it follows that
$$\tan\alpha = \mathbf{I} - \sqrt{\mathbf{J}}, \quad \tan\beta = \mathbf{K}.$$