Sine Rule

We are given $$AB = 5$$, $$\angle B = \cos^{-1}\!\left(\frac{3}{5}\right)$$ (so $$\cos B = \frac{3}{5}$$ and $$\sin B = \frac{4}{5}$$), and circumradius $$R = 5$$.

By the law of sines, the side opposite to $$B$$ (i.e., $$AC$$) satisfies: $$\frac{AC}{\sin B} = 2R \implies AC = 2 \times 5 \times \frac{4}{5} = 8.$$

Now applying the law of cosines with $$BC = a$$: $$AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos B$$ $$64 = 25 + a^2 - 2(5)(a)\!\left(\frac{3}{5}\right) = 25 + a^2 - 6a.$$

This gives $$a^2 - 6a - 39 = 0$$, so $$a = \frac{6 \pm \sqrt{36 + 156}}{2} = \frac{6 \pm \sqrt{192}}{2} = 3 \pm 4\sqrt{3}.$$

Since $$a = BC > 0$$, we have $$BC = 3 + 4\sqrt{3}$$.

The area of triangle $$ABC$$ is: $$\text{Area} = \frac{1}{2} \cdot AB \cdot BC \cdot \sin B = \frac{1}{2} \cdot 5 \cdot (3 + 4\sqrt{3}) \cdot \frac{4}{5} = 2(3 + 4\sqrt{3}) = 6 + 8\sqrt{3}.$$

1. Use the Fundamental Tangent Identity for Triangles

For any non-right-angled triangle $$\Delta ABC$$, the sum of the tangents of the angles is equal to their product:

$$\tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C$$

Let us substitute this shared value with a single variable, $$x$$:

$$x = \tan A + \tan B + \tan C = \tan A \cdot \tan B \cdot \tan C$$

Since the triangle is given as non-obtuse (and tangents are well-defined, meaning it is strictly acute), all individual tangent values are positive. Therefore, $$x > 0$$.

2. Analyze the Boundary Condition

For any acute triangle, the minimum value of the sum of tangents occurs when the triangle is equilateral ($$A = B = C = 60^\circ$$). This gives the standard geometric inequality:

$$\tan A + \tan B + \tan C \geq 3\sqrt{3}$$

$$x \geq 3\sqrt{3}$$

Substituting $$x$$ into the given equation:

$$\log_3(x) = \frac{1}{2\sqrt{3}}x$$

*(Note: The coefficient $$\frac{2}{3\sqrt{3}}$$ in standard versions of this problem is structured around the unique boundary condition of the function $$\frac{\log_3 x}{x}$$, which reaches its maximum valid trigonometric threshold precisely at $$x = 3\sqrt{3}$$, forcing equality).*

Let us check the value at the minimum boundary $$x = 3\sqrt{3}$$:

$$\text{LHS} = \log_3(3\sqrt{3}) = \log_3(3^{3/2}) = \frac{3}{2}$$

$$\text{RHS} = \frac{1}{2\sqrt{3}}(3\sqrt{3}) = \frac{3}{2}$$

Because the equation holds strictly at the minimum possible boundary condition, the triangle must be **equilateral**. Thus:

$$A = B = C = 60^\circ$$

3. Calculate the Side Length

The relationship between a side $$a$$ of an equilateral triangle and its circumradius $$R$$ is given by the law of sines:

$$a = 2R \sin A$$

Given that the circumradius $$R = 4$$:

$$a = 2(4) \sin(60^\circ) = 8 \cdot \frac{\sqrt{3}}{2} = 4\sqrt{3}$$

4. Compute the Area of the Triangle

The area of an equilateral triangle with side length $$a$$ is given by the formula:

$$\text{Area} = \frac{\sqrt{3}}{4}a^2$$

Substitute $$a = 4\sqrt{3}$$ into the area formula:

$$\text{Area} = \frac{\sqrt{3}}{4}(4\sqrt{3})^2$$

$$\text{Area} = \frac{\sqrt{3}}{4}(48) = 12\sqrt{3}$$