Let sin A / sin B = sin(A-C) / sin(C-B), where A, B, C are angles of a triangle ABC. If the lengths

Q: Wait, why can't we just use the Sine Rule at the very beginning? Why do we need to do all this trigonometric manipulation first?

Great question! The Sine Rule works on individual angles like $\sin A$ or $\sin B$. But our equation has terms like $\sin(A-C)$. You cannot simply replace $\sin(A-C)$ with $\frac{a-c}{2R}$. That is a common trap! We must simplify the expression into individual angles first to make the Sine Rule applicable.

Q: How do I know if $b^2, c^2, a^2$ are in A.P. or $a^2, b^2, c^2$ are in A.P.? I always get the order wrong.

Patience, my friend. The condition for three numbers $x, y, z$ to be in A.P. is $2y = x + z$. In our result, we found $2c^2 = a^2 + b^2$. This means $c^2$ is the middle term. So, $a^2, c^2, b^2$ (or $b^2, c^2, a^2$) are in A.P. Always look at which variable is doubled—that is your middle term!

Q: Is there a geometric way to solve this without all the algebra?

While geometry is beautiful, sometimes the algebraic path is the most direct. You could try to construct a triangle, but the trigonometric identity approach is the 'JEE way'—it's fast, elegant, and less prone to drawing errors. Master the algebra, and the geometry will follow.

Question

Let $\frac{\sin A}{\sin B} = \frac{\sin(A-C)}{\sin(C-B)}$, where $A, B, C$ are angles of a triangle $ABC$. If the lengths of the sides opposite these angles are $a, b, c$ respectively, then :

Options:

(A) $b^2 - a^2 = a^2 + c^2$

(B) $b^2, c^2, a^2$ are in A.P.

(C) $c^2, a^2, b^2$ are in A.P.

(D) $a^2, b^2, c^2$ are in A.P.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, open field, holding a triangle. You know the angles, but you want to know the sides. This is the classic struggle of trigonometry—the bridge between the abstract world of angles and the physical world of lengths.

We start with the given equation:
$$\frac{\sin A}{\sin B} = \frac{\sin(A-C)}{\sin(C-B)}$$

At first glance, it looks intimidating. We have composite angles like $(A-C)$ and $(C-B)$. But remember the most fundamental property of any triangle: $A + B + C = \pi$.

### The Angle Problem

This property is our secret weapon. We can rewrite $A$ as $\pi - (B+C)$ and $B$ as $\pi - (A+C)$. Because $\sin(\pi - 	heta) = \sin 	heta$, we can simplify the terms.

Suddenly, $\sin A$ becomes $\sin(B+C)$ and $\sin B$ becomes $\sin(A+C)$. We have transformed the equation into:
$$\frac{\sin(B+C)}{\sin(A+C)} = \frac{\sin(A-C)}{\sin(C-B)}$$

Look at the symmetry! We are now ready to proceed with the algebraic manipulation.

### The Algebraic Dance

Now, we cross-multiply. This is where the magic happens:
$$\sin(B+C)\sin(C-B) = \sin(A+C)\sin(A-C)$$

Do you see the pattern? This follows the classic identity $\sin(x+y)\sin(x-y) = \sin^2 x - \sin^2 y$. Applying this, the left side collapses into $\sin^2 C - \sin^2 B$, and the right side becomes $\sin^2 A - \sin^2 C$.

Our complex equation has simplified into a beautiful relationship of squares:
$$\sin^2 C - \sin^2 B = \sin^2 A - \sin^2 C$$

Rearranging this, we obtain:
$$2\sin^2 C = \sin^2 A + \sin^2 B$$

### The Bridge: The Sine Rule

We have a relationship of angles, but the question asks for sides. How do we bridge this gap? The Sine Rule is our bridge:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$

This implies $\sin A = \frac{a}{2R}$, $\sin B = \frac{b}{2R}$, and $\sin C = \frac{c}{2R}$. Let's substitute these into our equation:
$$2\left(\frac{c}{2R}ight)^2 = \left(\frac{a}{2R}ight)^2 + \left(\frac{b}{2R}ight)^2$$

The denominators of $4R^2$ are present in every term. They cancel out instantly, leaving us with the elegant result:
$$2c^2 = a^2 + b^2$$

### The Conclusion

Finally, we interpret the result. The condition for three numbers $x, y, z$ to be in Arithmetic Progression (A.P.) is $2y = x + z$.

Here, we have $2c^2 = a^2 + b^2$. This means $c^2$ is the arithmetic mean of $a^2$ and $b^2$.

Therefore, **$a^2, c^2, b^2$ are in A.P.** You have successfully navigated the trap, used the identity, and bridged the gap between trigonometry and geometry. That is the JEE spirit!

Animated Solution for Mathematics - Trigonometry: Let $\frac{s i n A}{s i n B} = \frac{s i n ( A - C )}{s i n ( C - B )}$ , where $A, B, C$ are angles of a triangle $A B C$ . If the lengths of the sides opposite these angles are $a, b, c$ respectively, then :

Select Answer:

Visualized Solution

The Given Trigonometric Equation

Angle Sum Property

Transforming $sin B$

Substituting into the Equation

Cross-Multiplication

The Key Trigonometric Identity

Applying the Identity

Grouping Sine Terms

Introducing the Sine Rule

Substituting Side Lengths

Simplifying the Equation

Arithmetic Progression Check

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Angle Problem

The Algebraic Dance

The Bridge

The Conclusion

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