Match the following:
Column I
(A) In a triangle ΔXYZ, let a, b, and c be the lengths of the sides op

Q: Why can we just cancel $\sin Z$ in Part A?

Because $Z$ is an angle of a triangle, $\sin Z$ is never zero. It is a safe and elegant simplification that allows us to isolate the ratio $\lambda$.

Q: Why did we reject the factor $(2\sin Y + \sin X)$ in Part B?

In any triangle, angles $X$ and $Y$ are between $0$ and $\pi$, so $\sin X$ and $\sin Y$ are strictly positive. Their sum cannot be zero, so we can safely discard that factor.

Q: How did you know the bisector is $y=x$ in Part C?

The vectors $\vec{OX}$ and $\vec{OY}$ have equal magnitudes. The angle bisector of two vectors of equal length is simply their sum, which points along the line $y=x$.

Question

Match the following:

List-I:

P. In a triangle $\Delta XYZ$, let $a, b$, and $c$ be the lengths of the sides opposite to the angles $X, Y$ and $Z$, respectively. If $2(a^2 - b^2) = c^2$ and $\lambda = \frac{\sin(X - Y)}{\sin Z}$, then possible values of $n$ for which $\cos(n\pi\lambda) = 0$ is (are)

Q. In a triangle $\Delta XYZ$, let $a, b$ and $c$ be the lengths of the sides opposite to the angles $X, Y$, and $Z$ respectively. If $1 + \cos 2X - 2\cos 2Y = 2\sin X \sin Y$, then possible value(s) of $\frac{a}{b}$ is (are)

R. In $R^2$, let $\sqrt{3}\hat{i} + \hat{j}, \hat{i} + \sqrt{3}\hat{j}$ and $\beta\hat{i} + (1 - \beta)\hat{j}$ be the position vectors of $X, Y$ and $Z$ with respect to the origin $O$, respectively. If the distance of $Z$ from the bisector of the acute angle of $\vec{OX}$ with $\vec{OY}$ is $\frac{3}{\sqrt{2}}$, then possible value(s) of $|\beta|$ is (are)

S. Suppose that $F(\alpha)$ denotes the area of the region bounded by $x = 0, x = 2, y^2 = 4x$ and $y = |\alpha x - 1| + |\alpha x - 2| + \alpha x$, where $\alpha \in \{0, 1\}$. Then the value(s) of $F(\alpha) + \frac{8}{3}\sqrt{2}$, when $\alpha = 0$ and $\alpha = 1$, is (are)

List-II:

1. 1

2. 2

3. 3

4. 5

5. 6

Sigma Percentile · Accepted Answer

### Analyzing the Geometric Relation

We begin with the given relation $2(a^2 - b^2) = c^2$. When dealing with side lengths in a triangle, the most effective strategy is to apply the Sine Rule: $a = 2R \sin X$, $b = 2R \sin Y$, and $c = 2R \sin Z$.

Substituting these into the original equation, the circumradius $R$ cancels out, yielding:

$$2(\sin^2 X - \sin^2 Y) = \sin^2 Z$$

Using the trigonometric identity $\sin^2 X - \sin^2 Y = \sin(X-Y)\sin(X+Y)$ and noting that $\sin(X+Y) = \sin Z$, we simplify the expression to:

$$2\sin(X-Y)\sin Z = \sin^2 Z$$

Since $\sin Z 
eq 0$ in any triangle, we divide both sides to find the ratio:

$$\lambda = \frac{\sin(X-Y)}{\sin Z} = \frac{1}{2}$$

Solving $\cos(n\pi\lambda) = 0$ now reduces to identifying the odd multiples of $\frac{\pi}{2}$.

### The Quadratic Trap

Moving to the second part, we analyze the equation $1 + \cos 2X - 2\cos 2Y = 2\sin X \sin Y$. We convert the double angles into sine-squared terms to unify the expression:

$$1 + (1 - 2\sin^2 X) - 2(1 - 2\sin^2 Y) = 2\sin X \sin Y$$

Simplifying the terms leads to the following quadratic equation:

$$2\sin^2 Y - \sin^2 X - \sin X \sin Y = 0$$

Factoring this expression yields:

$$(2\sin Y + \sin X)(\sin Y - \sin X) = 0$$

Because the first factor is strictly positive for any triangle, we are forced to conclude that $\sin Y = \sin X$. Consequently, $X = Y$, which implies the ratio $\frac{a}{b} = 1$.

### The Elegance of Vectors

We are given vectors $\vec{OX} = \sqrt{3}\hat{i} + \hat{j}$ and $\vec{OY} = \hat{i} + \sqrt{3}\hat{j}$. Note that their magnitudes are equal:

$$|\vec{OX}| = |\vec{OY}| = \sqrt{(\sqrt{3})^2 + 1^2} = 2$$

Since the vectors have equal magnitude, their angle bisector lies along the line $y=x$. The distance from a point $Z(\beta, 1-\beta)$ to the line $x-y=0$ is given by:

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

Solving $|2\beta - 1| = 3$ results in $\beta = 2$ or $\beta = -1$. Thus, the possible values for $|\beta|$ are **2** or **1**.

### The Piecewise Dance

Finally, we calculate the area under the curve. For $\alpha=0$, the function is $y=3$, and the area is:

$$\int_{0}^{2} (3 - 2\sqrt{x}) \, dx$$

For $\alpha=1$, the function $y = |x-1| + |x-2| + x$ changes behavior at $x=1$. We split the integral at this critical point to evaluate the area.

The beauty of this problem lies in the final cancellation: the irrational terms involving $\sqrt{2}$ vanish. This leaves us with clean, integer results, confirming the **symmetry and consistency** of the entire derivation.

Animated Solution for Mathematics - Vector Algebra: Match the following:

List-I

List-II

Select Matching Pairs:

Visualized Solution

Part A: Sine Rule Application

Part A: Trigonometric Simplification

Part A: Solving for $λ$ and $n$

Part B: Trigonometric Ratio Setup

Part B: Factorization

Part B: Conclusion

Part C: Vectors and Bisector

Part C: Distance Formula

Part C: Solving for $β$

Part D: Area for $α = 0$

Part D: Compute Area $F (0)$

Part D: Area for $α = 1$

Part D: Compute Area $F (1)$

The Sigma Insight: Components of a Vector

Analyzing the Geometric Relation

The Quadratic Trap

The Elegance of Vectors

The Piecewise Dance

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