Animated Solution for Mathematics - Trigonometry: 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 $\text{gcd}(m,n)=1$, then $m+n$ is equal to

Welcome, fellow traveler on the JEE Advanced path. Today, we are not just solving a problem; we are uncovering the hidden geometry of a triangle.

Imagine you are standing in a field, holding a compass and a ruler. You are given two sides of a triangle, $a=7$ and $b=8$, and the cosine of the angle between them, $\cos A=\frac{2}{3}$.

The third side, $c$, is hidden behind the variable $\alpha$, which we are told must be a natural number. In the world of competitive exams, a constraint like $\alpha \in \mathbb{N}$ is a neon sign pointing toward a specific, integer-based solution.

Our Geometric Compass

To find $\alpha$, we need a bridge between the sides and the angle. That bridge is the Law of Cosines.

We write it down:

Substituting our known values, we get:

As we simplify the squares, $8^2$ becomes $64$ and $7^2$ becomes $49$. The numerator simplifies to ${\alpha }^2+15$.

Now, we cross-multiply: $2\times 16\alpha =3\times ({\alpha }^2+15)$, which leads us to the quadratic equation:

Solving this is like solving a puzzle; we split the middle term to get $(3\alpha -5)(\alpha -9)=0$. We find two roots: $\alpha =9$ and $\alpha =\frac{5}{3}$.

But remember our constraint? $\alpha$ must be a natural number. Thus, the fraction is rejected, and our triangle is revealed: sides $7,8,$ and $9$.

Now that we know the sides, we need to evaluate $49\cos (3C)+42$. To get to $\cos (3C)$, we first need $\cos C$.

We return to the Cosine Rule:

Plugging in our sides $7,8,$ and $9$, we calculate:

Now, we invoke the triple angle identity: $\cos (3C)=4{\cos }^{3}C-3\cos C$. This is where the magic happens.

We substitute $\cos C=\frac{2}{7}$ into the identity:

Finally, we tackle the expression $49\cos (3C)+42$. Substituting our value for $\cos (3C)$, we have:

Distributing the $49$ is the key. $49\times \frac{32}{343}$ simplifies to $\frac{32}{7}$ because $343=49\times 7$.

Then, $49\times \frac{6}{7}$ becomes $7\times 6=42$. Our expression becomes $\frac{32}{7}-42+42$.

The $-42$ and $+42$ cancel out with elegant precision, leaving us with $\frac{32}{7}$. The problem asks for $m+n$ where $\frac{m}{n}=\frac{32}{7}$.

Since $\text{gcd}(32,7)=1$, we have $m=32$ and $n=7$. Adding them gives 39.