In a triangle ABC, if cos A + 2 cos B + cos C = 2 and the lengths of the sides opposite to the angle

Q: Why can't we just use the Sine Rule? It feels like it should be easier.

That is a great intuition, but look at the structure of the problem. The Sine Rule relates sides to the *sines* of the angles. Our given equation is locked entirely in the world of *cosines*. While you could technically convert sines to cosines, it would introduce square roots and make the algebra significantly more complex. The Cosine Rule is the direct, elegant path that keeps the algebra clean and manageable.

Q: I solved the cubic equation and got $b=4$ as a root, but the solution rejects it. Why?

This is the classic 'JEE Trap'! Always remember the Triangle Inequality: the sum of any two sides must be *strictly* greater than the third side ($a + b > c$). If $b=4$, then $3 + 4 = 7$, which is not strictly greater than $7$. This would result in a 'degenerate' triangle—a flat line. Since we need a valid triangle, we must reject $b=4$ and look for the value that satisfies the inequality.

Q: Wait, I calculated $\cos C$ and got a negative number. Did I make a mistake?

Not at all! A negative cosine value is perfectly valid in trigonometry. It simply tells us that the angle $C$ is obtuse (greater than $90^\circ$). In the context of this triangle, it just means the side $c$ is long enough to push the angle past the right-angle mark. Trust your algebra; the math is telling you the truth about the geometry.

Question

In a triangle $ABC$, if $\cos A + 2 \cos B + \cos C = 2$ and the lengths of the sides opposite to the angles $A$ and $C$ are 3 and 7 respectively, then $\cos A - \cos C$ is equal to

Options:

(A) $\frac{9}{7}$

(B) $\frac{10}{7}$

(C) $\frac{5}{7}$

(D) $\frac{3}{7}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the JEE journey. Today, we are not just solving a problem; we are uncovering the hidden architecture of a triangle. When you look at a problem like this, it is easy to feel overwhelmed by the variables.

We have $\cos A + 2 \cos B + \cos C = 2$, and we are given two sides, $a=3$ and $c=7$. It feels like a puzzle with missing pieces. But I want you to take a deep breath. In geometry, every equation is a constraint, and every constraint is a clue.

### The Bridge Between Worlds

We are given information about angles (via their cosines) and information about sides. How do we connect these two worlds? This is where the Cosine Rule becomes our most trusted ally.

Recall the standard form:
$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$

Why is this so powerful? Because in our problem, $a$ and $c$ are fixed constants ($3$ and $7$). The only variable that determines the 'personality' of this triangle is the side $b$. By expressing $\cos A$, $\cos B$, and $\cos C$ in terms of $b$, we transform a trigonometric equation into a purely algebraic one.

### The Algebraic Transformation

Let us perform the substitution. We take our given equation, $\cos A + 2 \cos B + \cos C = 2$, and replace the terms.

For $\cos A$, we have:
$$\frac{b^2 + 7^2 - 3^2}{2(b)(7)} = \frac{b^2 + 40}{14b}$$

For $\cos B$, we have:
$$\frac{3^2 + 7^2 - b^2}{2(3)(7)} = \frac{58 - b^2}{42}$$

For $\cos C$, we have:
$$\frac{3^2 + b^2 - 7^2}{2(3)(b)} = \frac{b^2 - 40}{6b}$$

Now, look at the equation:
$$\frac{b^2 + 40}{14b} + 2\left(\frac{58 - b^2}{42}\right) + \frac{b^2 - 40}{6b} = 2$$

I know, it looks like a mess of fractions. But do not let the denominators intimidate you. We find the Least Common Multiple of $14b$, $42$, and $6b$, which is $42b$. Multiplying the entire equation by $42b$ clears the fractions and leaves us with a clean, manageable polynomial.

### The Cubic Challenge

After multiplying by $42b$ and expanding the terms, we arrive at:
$$3(b^2 + 40) + 2b(58 - b^2) + 7(b^2 - 40) = 84b$$

Expanding this, we get $3b^2 + 120 + 116b - 2b^3 + 7b^2 - 280 = 84b$. Grouping the terms, we find ourselves staring at a cubic equation:
$$-2b^3 + 10b^2 + 32b - 160 = 0$$

Divide by $-2$, and we get $b^3 - 5b^2 - 16b + 80 = 0$. Using factoring by grouping:
$$b^2(b - 5) - 16(b - 5) = 0 \Rightarrow (b - 5)(b^2 - 16) = 0$$

The roots are $b=5$, $b=4$, and $b=-4$.

### The Reality Check

Here is where the true test of a mathematician lies. We have three mathematical solutions, but do they all exist in the physical world?

1. $b = -4$: A side length cannot be negative. We discard this immediately.
2. $b = 4$: We must check the Triangle Inequality. If $b=4$, then $a+b = 3+4 = 7$. Since $a+b$ is not strictly greater than $c$ ($7$), this triangle cannot exist.
3. $b = 5$: Here, $3+5 > 7$. This works perfectly!

So, our triangle is defined by sides $3, 5,$ and $7$.

### The Final Victory

With $b=5$ locked in, the rest is just a victory lap. We calculate:
$$\cos A = \frac{5^2 + 7^2 - 3^2}{2(5)(7)} = \frac{65}{70} = \frac{13}{14}$$

Similarly:
$$\cos C = \frac{3^2 + 5^2 - 7^2}{2(3)(5)} = \frac{-15}{30} = -\frac{1}{2}$$

The question asks for $\cos A - \cos C$:
$$\frac{13}{14} - \left(-\frac{1}{2}\right) = \frac{13}{14} + \frac{7}{14} = \frac{20}{14} = \frac{10}{7}$$

There it is. The elegance of the final answer, **$\frac{10}{7}$**, is the reward for your persistence. You navigated the constraints, avoided the traps, and arrived at the truth.

Animated Solution for Mathematics - Trigonometry: In a triangle $A B C$ , if $cos A + 2 cos B + cos C = 2$ and the lengths of the sides opposite to the angles $A$ and $C$ are 3 and 7 respectively, then $cos A - cos C$ is equal to

Select Answer:

Visualized Solution

Identifying the Given Information

Applying the Cosine Rule

Substituting Values into the Equation

Simplifying the Numerators

Clearing the Denominators

Expanding and Rearranging

Solving the Cubic Equation

Applying Triangle Inequality

Calculating $cos A$ and $cos C$

Final Result

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Bridge Between Worlds

The Algebraic Transformation

The Cubic Challenge

The Reality Check

The Final Victory

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