If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the

Q: Wait, why do we reject the negative root here? Or rather, why can't we just assume it's an equilateral triangle?

That is a fantastic question! While it is true that if $A=B=C=60^\circ$, the condition of A.P. is satisfied, we cannot *assume* the triangle is equilateral. The A.P. condition only fixes the middle angle $B$ to be $60^\circ$. The other two angles $A$ and $C$ could be anything as long as $A+C = 120^\circ$. The beauty of this problem is that the expression evaluates to the same value regardless of the specific values of $A$ and $C$!

Q: Do I really need the Sine Rule here, or can I use the Cosine Rule?

You *could* use the Cosine Rule, but it would make the algebra significantly more complex. The Sine Rule is the perfect tool here because it directly relates the sides to the sines of the angles, which matches the trigonometric functions in your expression. Always look for the path of least resistance in JEE problems!

Q: What if the angles weren't in A.P.? Would the expression still simplify?

If the angles were not in A.P., the expression would depend on the specific values of $A, B,$ and $C$. The A.P. condition is the 'key' that unlocks the simplification by fixing $B$ and allowing us to use the identity $\sin(A+C) = \sin(180^\circ - B) = \sin B$. Without that constraint, the expression would not collapse so elegantly.

Question

If the angles $A, B$ and $C$ of a triangle are in an arithmetic progression and if $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively, then the value of the expression $\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$ is

Options:

(A) $1/2$

(B) $\sqrt{3}/2$

(C) 1

(D) $\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a trigonometry problem; we are uncovering a hidden symmetry.

Imagine you are standing in a field, holding a triangle $ABC$. You know nothing about its specific shape, only that its angles $A, B,$ and $C$ are dancing in an arithmetic progression. This simple constraint is the key to unlocking a beautiful, elegant result.

### The $60^\circ$ Revelation

Let us start with the foundation. We are told the angles $A, B,$ and $C$ are in an arithmetic progression. In the language of algebra, this means the middle term $B$ is the average of the other two: $2B = A + C$.

Now, we invoke the universal law of triangles: the sum of interior angles is always $180^\circ$. So, $A + B + C = 180^\circ$.

By substituting our A.P. condition ($A+C = 2B$) into this sum, we get $2B + B = 180^\circ$, which simplifies to $3B = 180^\circ$. Just like that, we have discovered that $B = 60^\circ$. This is our anchor point.

### The Bridge of Sine Rule

Now, look at the expression we need to evaluate:

$$\frac{a}{c} \sin 2C + \frac{c}{a} \sin 2A$$

It is a mix of side lengths and trigonometric functions. How do we bridge this gap? We use the Sine Rule, the great translator of geometry.

The Sine Rule tells us that $\frac{a}{\sin A} = \frac{c}{\sin C} = k$. This allows us to replace the side lengths $a$ and $c$ with $k \sin A$ and $k \sin C$.

When we substitute these into our expression, the constant $k$ cancels out entirely, leaving us with a pure trigonometric expression:

$$\frac{\sin A}{\sin C} \sin 2C + \frac{\sin C}{\sin A} \sin 2A$$

### The Algebraic Choreography

This is where the magic happens. We have double angles, $\sin 2C$ and $\sin 2A$, which look intimidating. But remember your toolkit: the double-angle identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$.

Let us apply this to our expression:

$$\frac{\sin A}{\sin C} (2 \sin C \cos C) + \frac{\sin C}{\sin A} (2 \sin A \cos A)$$

Watch closely as the denominators vanish. The $\sin C$ in the first term cancels, and the $\sin A$ in the second term cancels. We are left with $2 \sin A \cos C + 2 \sin C \cos A$.

Factoring out the $2$, we get $2(\sin A \cos C + \cos A \sin C)$.

### The Final Flourish

Does that bracket look familiar? It is the classic compound angle identity for $\sin(A+C)$.

So, our expression is simply $2 \sin(A+C)$. Since $A+C = 180^\circ - B$, we have $2 \sin(180^\circ - B)$, which is $2 \sin B$.

We already know $B = 60^\circ$, so the final answer is:

$$2 \sin 60^\circ = 2 \left( \frac{\sqrt{3}}{2} ight) = \mathbf{\sqrt{3}}$$

Isn't it satisfying? We started with a complex expression and, through the logic of A.P. and the elegance of trigonometric identities, we arrived at a simple, clean result. Keep this mindset—look for the hidden structure, and the math will always reveal its secrets.

Animated Solution for Mathematics - Trigonometry: If the angles $A, B$ and $C$ of a triangle are in an arithmetic progression and if $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively, then the value of the expression $\frac{a}{c} sin 2 C + \frac{c}{a} sin 2 A$ is

Select Answer:

Visualized Solution

Visualizing Triangle $A B C$

The Arithmetic Progression Condition

Angle Sum Property

Calculating Angle $B$

The Sine Rule

Substituting into the Expression

Simplifying the Fractions

Double Angle Formula

Expanding the Double Angles

Canceling Denominators

Factoring the Expression

Compound Angle Identity

Substituting $A + C$

Final Evaluation

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The $6 0^{\circ}$ Revelation

The Bridge of Sine Rule

The Algebraic Choreography

The Final Flourish

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