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

Given a triangle \(ABC\) with side lengths \(BC=7\) and \(AC=8\), and the cosine of the included angle \(\cos A = \frac{2}{3}\). The third side \(AB\), denoted as \(\alpha\), is a natural number. The objective is to determine the value of \(m+n\), where \(\frac{m}{n}\) represents the simplified fractional value of \( 49 \cos(3C) + 42 \).

Concept Used:

This solution employs fundamental trigonometric laws for triangles and a key trigonometric identity.

1. Law of Cosines: For a triangle with sides \(a, b, c\) and opposite angles \(A, B, C\), the following equations apply: \[ a^2 = b^2 + c^2 - 2bc \cos A \] \[ c^2 = a^2 + b^2 - 2ab \cos C \]
2. Triple Angle Identity for Cosine: The cosine of a triple angle can be expressed as: \[ \cos(3\theta) = 4\cos^3(\theta) - 3\cos(\theta) \]

Step-by-Step Solution:

Step 1: Determine the length of side \(AB = \alpha\).

We have sides \(a = BC = 7\), \(b = AC = 8\), and \(c = AB = \alpha\). The angle \(A\) has \(\cos A = \frac{2}{3}\). Applying the Law of Cosines for angle \(A\):

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

Substitute the given values:

\[7^2 = 8^2 + \alpha^2 - 2(8)(\alpha)\left(\frac{2}{3}\right)\]\[49 = 64 + \alpha^2 - \frac{32\alpha}{3}\]

Rearrange into a quadratic equation in \(\alpha\):

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

Multiply by 3 to eliminate the fraction:

\[3\alpha^2 - 32\alpha + 45 = 0\]

Factor the quadratic equation:

\[(3\alpha - 5)(\alpha - 9) = 0\]

The possible values for \(\alpha\) are \(\alpha = \frac{5}{3}\) and \(\alpha = 9\). Since \(\alpha\) must be a natural number (\(\alpha \in \mathbb{N}\)), we select \(\alpha = 9\). Thus, the length of side \(AB\) is 9.

Step 2: Calculate \(\cos C\).

With all three sides known (\(a=7, b=8, c=9\)), we use the Law of Cosines to find \(\cos C\):

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

Substitute the side lengths:

\[9^2 = 7^2 + 8^2 - 2(7)(8) \cos C\]\[81 = 49 + 64 - 112 \cos C\]\[81 = 113 - 112 \cos C\]\[112 \cos C = 113 - 81 = 32\]\[\cos C = \frac{32}{112} = \frac{2}{7}\]

Step 3: Compute \(\cos(3C)\).

Using the triple angle identity \(\cos(3C) = 4\cos^3(C) - 3\cos(C)\) with \(\cos C = \frac{2}{7}\):

\[\cos(3C) = 4\left(\frac{2}{7}\right)^3 - 3\left(\frac{2}{7}\right)\]\[\cos(3C) = 4\left(\frac{8}{343}\right) - \frac{6}{7}\]\[\cos(3C) = \frac{32}{343} - \frac{294}{343}\]\[\cos(3C) = -\frac{262}{343}\]

Step 4: Evaluate the expression \( 49 \cos(3C) + 42 \).

\[49 \left(-\frac{262}{343}\right) + 42\]

Simplify using \(343 = 49 \times 7\):

\[-\frac{262}{7} + 42\]

Combine the terms with a common denominator:

\[-\frac{262}{7} + \frac{294}{7} = \frac{32}{7}\]

Step 5: Find the value of \(m+n\).

The evaluated expression is \(\frac{32}{7}\). Given this equals \(\frac{m}{n}\) in simplified form, we have \(m=32\) and \(n=7\). Their greatest common divisor is 1, confirming simplification.

The required sum is \(m+n\):

\[m + n = 32 + 7 = 39\]

The final answer is 39.