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

The problem provides the lengths of two sides of a triangle \(ABC\) as \(BC=7\) and \(AC=8\), and the cosine of the included angle \(\cos A = \frac{2}{3}\). The third side \(AB = \alpha\) is a natural number. We are asked to find the value of \(m+n\), where \( \frac{m}{n} \) is the simplified fractional value of the expression \( 49 \cos(3C) + 42 \).

Concept Used:

The solution requires the application of fundamental trigonometric rules for a triangle and a trigonometric identity.

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

Step-by-Step Solution:

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

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

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

Substituting the given values:

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

Rearranging the terms to form a quadratic equation in \(\alpha\):

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

Multiplying by 3 to clear the fraction:

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

We solve this quadratic equation for \(\alpha\):

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

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

Step 2: Calculate the value of \(\cos C\).

Now that we know all three sides of the triangle (\(a=7, b=8, c=9\)), we can use the Law of Cosines again to find \(\cos C\):

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

Substituting the side lengths:

\[ 9^2 = 7^2 + 8^2 - 2(7)(8) \cos C \] \[ 81 = 49 + 64 - 112 \cos C \] \[ 81 = 113 - 112 \cos C \] \[ 112 \cos C = 113 - 81 = 32 \] \[ \cos C = \frac{32}{112} = \frac{2 \times 16}{7 \times 16} = \frac{2}{7} \]

Step 3: Compute the value of \(\cos(3C)\).

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

\[ \cos(3C) = 4\left(\frac{2}{7}\right)^3 - 3\left(\frac{2}{7}\right) \] \[ \cos(3C) = 4\left(\frac{8}{343}\right) - \frac{6}{7} \] \[ \cos(3C) = \frac{32}{343} - \frac{6 \times 49}{7 \times 49} = \frac{32}{343} - \frac{294}{343} \] \[ \cos(3C) = \frac{32 - 294}{343} = -\frac{262}{343} \]

Step 4: Evaluate the given expression.

We need to find the value of \( 49 \cos(3C) + 42 \).

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

Since \(343 = 7^3 = 49 \times 7\), we can simplify the expression:

\[ \frac{49}{49 \times 7} \times (-262) + 42 = -\frac{262}{7} + 42 \]

To combine these terms, we find a common denominator:

\[ -\frac{262}{7} + \frac{42 \times 7}{7} = \frac{-262 + 294}{7} = \frac{32}{7} \]

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

The expression evaluates to \(\frac{32}{7}\). We are given that this is equal to \(\frac{m}{n}\), where \(\gcd(m, n) = 1\). Here, \(m=32\) and \(n=7\). The greatest common divisor of 32 and 7 is 1. Therefore, the values are correct.

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

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

The final answer is 39.

Answer 2

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