In a triangle $$ABC$$, $$BC = 7$$, $$AC = 8$$, $$AB = \alpha \in \mathbb{N}$$ and $$\cos A = \frac{2}{3}$$. If $$49\cos(3C) + 42 = \frac{m}{n}$$, where $$\gcd(m, n) = 1$$, then $$m + n$$ is equal to ___________

In triangle $$ABC$$, $$BC = a = 7$$, $$AC = b = 8$$, $$AB = c = \alpha \in \mathbb{N}$$, and $$\cos A = 2/3$$.

Find $$\alpha$$ using the cosine rule.

$$a^2 = b^2 + c^2 - 2bc\cos A$$

$$49 = 64 + \alpha^2 - 2(8)(\alpha)(2/3) = 64 + \alpha^2 - \frac{32\alpha}{3}$$

$$\alpha^2 - \frac{32\alpha}{3} + 15 = 0$$

$$3\alpha^2 - 32\alpha + 45 = 0$$

$$\alpha = \frac{32 \pm \sqrt{1024 - 540}}{6} = \frac{32 \pm \sqrt{484}}{6} = \frac{32 \pm 22}{6}$$

$$\alpha = 9$$ or $$\alpha = 5/3$$ (not a natural number). So $$\alpha = 9$$.

Find $$\cos C$$ using the cosine rule.

$$c^2 = a^2 + b^2 - 2ab\cos C$$

$$81 = 49 + 64 - 2(7)(8)\cos C$$

$$81 = 113 - 112\cos C$$

$$\cos C = \frac{32}{112} = \frac{2}{7}$$

Compute $$\cos 3C$$.

$$\cos 3C = 4\cos^3 C - 3\cos C = 4\left(\frac{2}{7}\right)^3 - 3\left(\frac{2}{7}\right) = \frac{32}{343} - \frac{6}{7} = \frac{32 - 294}{343} = \frac{-262}{343}$$

Compute the expression.

$$49\cos 3C + 42 = 49 \cdot \frac{-262}{343} + 42 = \frac{-262}{7} + 42 = \frac{-262 + 294}{7} = \frac{32}{7}$$

So $$\frac{m}{n} = \frac{32}{7}$$ with $$\gcd(32, 7) = 1$$. Therefore $$m + n = 32 + 7 = 39$$.

The answer is 39.