Let ABC be a triangle such that ∠ACB = π/6 and let a, b and c denote the lengths of the sides opposi

Q: Why can't we use the Sine Rule here?

The Sine Rule requires knowing at least one side-angle pair (e.g., $\frac{a}{\sin A} = \frac{b}{\sin B}$). Here, we only have one angle and expressions for all three sides. The Law of Cosines is designed specifically to relate three sides and one angle, making it the perfect tool for this job.

Q: Why do we reject the negative root?

In the world of pure algebra, quadratic equations often yield two solutions. But in the world of geometry, side lengths must be positive. Our constraint $x > 1$ acts as a filter, ensuring the triangle is physically possible. The negative root leads to a negative side length, which is impossible in Euclidean geometry.

Q: How did you know to factor the numerator like that?

Experience! Expanding $(x^2 + x + 1)^2$ is a nightmare and a recipe for calculation errors. Whenever you see squares in a numerator like $a^2 + b^2 - c^2$, always look for the difference of squares $(a^2 - c^2)$ first. It is a classic JEE strategy to simplify complex expressions before doing the heavy lifting.

Question

Let $ABC$ be a triangle such that $\angle ACB = \pi/6$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value(s) of $x$ for which $a = x^2 + x + 1, b = x^2 - 1$ and $c = 2x + 1$ is (are)

Options:

(A) $-(2 + \sqrt{3})$

(B) $1 + \sqrt{3}$

(C) $2 + \sqrt{3}$

(D) $4	ext{√}3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are not just solving a problem; we are embarking on a journey. We have a triangle $ABC$ with a fixed angle $\angle C = \frac{\pi}{6}$ (or $30^\circ$), and its sides are defined by polynomials in $x$.

It looks like a daunting algebraic mess, but I want you to take a deep breath. In the JEE Advanced arena, the most complex-looking problems often hide the most elegant solutions. Let us decode this together.

### The Physical Reality

Before we touch a single equation, we must respect the physical reality of the triangle. A triangle is not just a set of variables; it is a shape that must exist in space. This means all side lengths must be strictly positive.

We are given $a = x^2 + x + 1$, $b = x^2 - 1$, and $c = 2x + 1$. For $b$ to be positive, $x^2 - 1 > 0$, which implies $x > 1$ or $x < -1$. For $c$ to be positive, $2x + 1 > 0$, meaning $x > -0.5$.

When we intersect these conditions, we find our **golden rule**: $x > 1$. This constraint is our compass; it will guide us when we reach the final step and have to choose between potential solutions.

### The Law of Cosines

Now, how do we bridge the gap between the sides and the angle? The Law of Cosines is our bridge. It states that:

$$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$

We know $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$. Substituting our expressions, we get:

$$\frac{\sqrt{3}}{2} = \frac{(x^2 + x + 1)^2 + (x^2 - 1)^2 - (2x + 1)^2}{2(x^2 + x + 1)(x^2 - 1)}$$

I know what you are thinking: "Do I really have to expand all these squares?" The answer is a resounding **NO**. In JEE, if you find yourself doing massive, tedious expansions, you are likely missing a shortcut.

### The Algebraic Dance

Let us look at the numerator: $(a^2 - c^2) + b^2$. This is the secret. We can write $a^2 - c^2$ as $(a - c)(a + c)$.

Let us calculate these:
$a - c = (x^2 + x + 1) - (2x + 1) = x^2 - x = x(x - 1)$
$a + c = (x^2 + x + 1) + (2x + 1) = x^2 + 3x + 2 = (x + 1)(x + 2)$

When we multiply these, we get $x(x - 1)(x + 1)(x + 2)$. Notice the magic? $(x - 1)(x + 1)$ is $x^2 - 1$. So, $a^2 - c^2 = (x^2 - 1)(x^2 + 2x)$.

Now, add $b^2 = (x^2 - 1)^2$ to this. We have a common factor of $(x^2 - 1)$! Factoring it out, we get:

$$(x^2 - 1)[(x^2 + 2x) + (x^2 - 1)] = (x^2 - 1)(2x^2 + 2x - 1)$$

The massive numerator has collapsed into a beautiful, manageable product.

### The Quadratic Finale

Now, we substitute this back into our equation. Since $x > 1$, we know $x^2 - 1 
eq 0$, so we can safely cancel it from the numerator and denominator. We are left with:

$$\sqrt{3}(x^2 + x + 1) = 2x^2 + 2x - 1$$

Rearranging this into a standard quadratic form, we get:

$$(2 - \sqrt{3})x^2 + (2 - \sqrt{3})x - (1 + \sqrt{3}) = 0$$

Dividing by $(2 - \sqrt{3})$ and rationalizing the constant term, we arrive at $x^2 + x - (5 + 3\sqrt{3}) = 0$. Applying the quadratic formula, we find the discriminant:

$$D = 1 + 4(5 + 3\sqrt{3}) = 21 + 12\sqrt{3}$$

Recognizing this as $(3 + 2\sqrt{3})^2$, we find the roots:

$$x = \frac{-1 \pm (3 + 2\sqrt{3})}{2}$$

The positive root is **$x = 1 + \sqrt{3}$**, and the negative root is rejected by our constraint $x > 1$. You have conquered the problem! Remember, the math is just the language; the logic is the art.

Animated Solution for Mathematics - Trigonometry: Let $A B C$ be a triangle such that $∠ A C B = π /6$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value(s) of $x$ for which $a = x^{2} + x + 1, b = x^{2} - 1$ and $c = 2 x + 1$ is (are)

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$

Establishing Side Constraints

Applying the Law of Cosines

The Raw Substitution

Smart Algebraic Factoring

Simplifying the Numerator

Total Numerator Expression

Canceling Common Factors

Forming the Quadratic Equation

Rationalizing the Constant Term

Applying the Quadratic Formula

Perfect Square in Discriminant

Final Result and Takeaway

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Physical Reality

The Law of Cosines

The Algebraic Dance

The Quadratic Finale

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