With usual notation, if in a triangle ABC; (b+c)/11 = (c+a)/12 = (a+b)/13, then prove that cos A / 7

Q: Why do we set the ratios to $k$? Can't we just solve them directly?

Setting the ratio to a constant $k$ is a classic technique in JEE problems involving proportions. It transforms a complex relational problem into a system of linear equations, making the variables $a$, $b$, and $c$ tangible. Without $k$, you are dealing with ratios; with $k$, you are dealing with actual lengths, which makes the subsequent algebra much cleaner.

Q: Why add all the equations together? Is there a faster way?

Adding the equations is the most elegant way to exploit the cyclic symmetry of the problem. If you try to substitute $b=11k-c$ into the other equations, you'll get there, but it's prone to algebraic errors. Summing them gives you $2(a+b+c) = 36k$ in one step—it's the 'master key' approach that simplifies the entire system.

Q: Do I really need the Cosine Rule? Can I use the Sine Rule?

The Sine Rule ($\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$) is powerful, but it gives you ratios of sines, not cosines. Since the problem asks for a relationship between $\cos A$, $\cos B$, and $\cos C$, the Cosine Rule is the direct path. It connects the side lengths directly to the cosine of the angle, which is exactly what the problem demands.

Question

With usual notation, if in a triangle $ABC$; $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$, then prove that $\frac{\cos A}{7} = \frac{\cos B}{19} = \frac{\cos C}{25}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are going to unravel the elegance of triangle geometry. When you look at a problem like this, it is easy to feel overwhelmed by the variables.

But look closer—there is a beautiful, rhythmic symmetry here. We are given:

$$\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$$

This is not just a set of equations; it is a pattern waiting to be unlocked. Our first step is to introduce a common constant, $k$.

By setting these ratios equal to $k$, we transform an abstract relationship into a concrete system:

$$b+c=11k, \quad c+a=12k, \quad a+b=13k$$

This is our foundation.

### The Master Key

Now, how do we solve this system efficiently? You could solve for one variable and substitute, but that is the long road. Instead, let's use the power of symmetry.

If we add all three equations together, we get:

$$(b+c) + (c+a) + (a+b) = 11k + 12k + 13k$$

Notice that each side, $a$, $b$, and $c$, appears exactly twice. This gives us $2(a+b+c) = 36k$.

Dividing by $2$, we find the 'master key':

$$a+b+c = 18k$$

This single equation is the key to everything. By subtracting our original equations from this sum, we can isolate $a$, $b$, and $c$ in terms of $k$:

$$a = 18k - 11k = 7k$$
$$b = 18k - 12k = 6k$$
$$c = 18k - 13k = 5k$$

We have successfully reduced the entire triangle to a simple set of side lengths proportional to $7$, $6$, and $5$.

### The Bridge to Angles

Now that we have the sides, we need to bridge the gap to the angles. This is where the Cosine Rule becomes our most trusted tool. The rule states that:

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

Let's apply this to angle $A$. Substituting our values, we get:

$$\cos A = \frac{(6k)^2 + (5k)^2 - (7k)^2}{2(6k)(5k)}$$

Notice how the $k^2$ terms in the numerator and denominator will cancel out? This is the beauty of the $k$ method!

After expanding, we get:

$$\cos A = \frac{36k^2 + 25k^2 - 49k^2}{60k^2} = \frac{12k^2}{60k^2} = \frac{1}{5}$$

We repeat this process for $\cos B$ and $\cos C$. For $\cos B$, we use $\frac{c^2 + a^2 - b^2}{2ca}$, which yields $\frac{19}{35}$. For $\cos C$, we use $\frac{a^2 + b^2 - c^2}{2ab}$, which yields $\frac{5}{7}$.

### The Final Victory

We are at the finish line. The problem asks us to prove that:

$$\frac{\cos A}{7} = \frac{\cos B}{19} = \frac{\cos C}{25}$$

Let's test our results:

$$\frac{\cos A}{7} = \frac{1/5}{7} = \frac{1}{35}$$
$$\frac{\cos B}{19} = \frac{19/35}{19} = \frac{1}{35}$$
$$\frac{\cos C}{25} = \frac{5/7}{25} = \frac{1}{35}$$

Every single ratio simplifies to $\frac{1}{35}$. The proof is complete!

It is a perfect, elegant result. Remember, in JEE Advanced, it is not just about the calculation; it is about seeing the structure. You have done well today!

Animated Solution for Mathematics - Trigonometry: With usual notation, if in a triangle $A B C$ ; $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ , then prove that $\frac{c o s A}{7} = \frac{c o s B}{19} = \frac{c o s C}{25}$ .

Visualized Solution

Triangle Setup

Setting Ratios to a Constant $k$

Forming Linear Equations

Summing the Equations

Finding the Perimeter Sum

Solving for Individual Sides

The Cosine Rule

Calculating $cos A$

Calculating $cos B$

Calculating $cos C$

Evaluating the Target Ratios

Verifying the Remaining Ratios

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Master Key

The Bridge to Angles

The Final Victory

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } With usual notation, if in a triangle ABC; 11b+c​=12c+a​=13a+b​, then prove that 7cosA​=19cosB​=25cosC​.