Question

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 ________.

Answer 1

Correct Answer: 39

Using the cosine rule for \( \cos A \):
\[\cos A = \frac{b^2 + c^2 - a^2}{2bc}\]
Substitute \( b = 8 \), \( c = 7 \), and \( \cos A = \frac{2}{3} \):
\[\frac{2}{3} = \frac{8^2 + 7^2 - a^2}{2 \times 8 \times 7}\]
\[\implies a^2 = 9\]
\[\implies a = 3\]
Now, calculate \( \cos C \) using the cosine rule:
\[\cos C = \frac{7^2 + 8^2 - 9^2}{2 \times 7 \times 8} = \frac{2}{7}\]
Then, for \( \cos(3C) \), we use the triple angle formula:
\[49 \cos(3C) + 42 = 49 \left( 4 \cos^3 C - 3 \cos C \right) + 42\]
Substituting \( \cos C = \frac{2}{7} \):
\[= 49 \left( 4 \left( \frac{2}{7} \right)^3 - 3 \cdot \frac{2}{7} \right) + 42\]
\[= 49 \left( \frac{32}{343} - \frac{6}{7} \right) + 42\]
\[= \frac{32}{7} + 42\]
Thus, \( m = 32 \) and \( n = 7 \), so:
\[m + n = 32 + 7 = 39\]