If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest,

Q: Wait, why do we reject the case where $a = b$? That would make it an isosceles triangle, which is a valid triangle, right?

That is a brilliant observation! Mathematically, if $a = b$, the equation $(a-b)(a^2 + ab - c^2) = 0$ is satisfied. However, look at the problem statement: the sides are in an Arithmetic Progression. If $a = b$, then for the sides to be in A.P., we would need $b = a$ and $c = a$ (since $2b = a+c$). This would make the triangle equilateral. But if the triangle were equilateral, all angles would be $60^\circ$, which contradicts the condition that the greatest angle is double the smallest ($C = 2A \implies 60^\circ = 120^\circ$, which is false). Thus, $a \neq b$ is a necessary constraint.

Q: Why can't we just use the Sine Rule for all three sides? Why bring in the Cosine Rule?

The Sine Rule is fantastic for relating sides to angles, but it only gives us ratios like $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$. While we can find $\frac{c}{a}$, we don't know $\sin B$ or $\sin C$ directly without more information. The Cosine Rule is the 'bridge' that allows us to convert the angular information ($\cos A$) back into the language of side lengths ($a, b, c$). It is the only way to close the loop and solve for the sides algebraically.

Q: Is there a shortcut? Can I just test the options provided?

In an exam setting, testing options is a valid strategy! If you check $4:5:6$, you can verify if the sides are in A.P. (they are) and if the angle condition holds. However, as a student of physics and mathematics, I encourage you to derive the result. The derivation teaches you how to manipulate trigonometric identities and algebraic polynomials—skills that are essential for the tougher, non-multiple-choice problems you will face in JEE Advanced.

Question

If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is :

Options:

(A) $5 : 9 : 13$

(B) $5 : 6 : 7$

(C) $4 : 5 : 6$

(D) $3 : 4 : 5$

Sigma Percentile · Accepted Answer

### The Geometry of Harmony: Unlocking the Triangle

Welcome, fellow traveler on this journey through the elegant world of trigonometry and algebra. Today, we are not just solving a problem; we are uncovering a hidden symmetry within a triangle.

Imagine you are standing in a field, holding three sticks of lengths $a, b,$ and $c$. You are told they form a triangle, they are in an Arithmetic Progression, and the largest angle is exactly twice the smallest.

### Phase 1: The Arithmetic Backbone

First, let us define our reality. We have sides $a < b < c$.

Because they are in an Arithmetic Progression, the middle term $b$ is the average of the extremes. This gives us our first anchor:

$$2b = a + c$$

This simple linear relationship is the foundation upon which we will build our entire solution. Keep this equation close; it will be the key to simplifying our final algebraic expression.

### Phase 2: The Angle Bridge

Now, consider the angles. We are told the greatest angle $C$ is double the smallest angle $A$. So, $C = 2A$.

We have a relationship between sides and a relationship between angles. How do we connect them? The Sine Rule is our bridge. It tells us that:

$$\frac{a}{\sin A} = \frac{c}{\sin C}$$

By rearranging this, we find that $\frac{c}{a} = \frac{\sin C}{\sin A}$. Substituting our angle condition, we get:

$$\frac{c}{a} = \frac{\sin 2A}{\sin A}$$

Here is where the magic happens. Using the double-angle identity, $\sin 2A = 2 \sin A \cos A$, the $\sin A$ terms cancel out beautifully, leaving us with:

$$\frac{c}{a} = 2 \cos A$$

We have successfully bridged the gap between the ratio of the sides and the cosine of the smallest angle.

### Phase 3: The Algebraic Clash

We are close, but we still have a trigonometric term, $\cos A$. We need to return to the world of pure algebra.

The Cosine Rule is our tool here. We know that:

$$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$$

Let us substitute this into our previous equation:

$$\frac{c}{a} = 2 \left( \frac{b^2 + c^2 - a^2}{2bc} ight)$$

The factor of $2$ cancels out, leaving us with:

$$\frac{c}{a} = \frac{b^2 + c^2 - a^2}{bc}$$

Now, cross-multiply to clear the fractions:

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

Expanding this, we get $bc^2 = ab^2 + ac^2 - a^3$. Rearranging everything to one side, we arrive at the polynomial:

$$a^3 - ab^2 + bc^2 - ac^2 = 0$$

This looks intimidating, but look closer. We can factor this! Grouping the terms gives us $a(a^2 - b^2) - c^2(a - b) = 0$. Using the difference of squares, we get:

$$(a - b)[a(a + b) - c^2] = 0$$

### Phase 4: The Final Resolution

We know $a 
eq b$ because the sides are in A.P. and distinct. Therefore, we must have $a^2 + ab - c^2 = 0$.

Now, we bring back our very first equation: $b = \frac{a+c}{2}$. Substituting this into our quadratic, we get:

$$a^2 + a\left(\frac{a+c}{2}ight) - c^2 = 0$$

Multiplying by $2$ to clear the fraction, we obtain $2a^2 + a^2 + ac - 2c^2 = 0$, which simplifies to:

$$3a^2 + ac - 2c^2 = 0$$

Factoring this quadratic, we find $(3a - 2c)(a + c) = 0$. Since the sum of sides $a+c$ cannot be zero, we are left with $3a = 2c$, or $c = \frac{3}{2}a$.

Substituting this back into our A.P. condition, we find $b = \frac{a + \frac{3}{2}a}{2} = \frac{5}{4}a$. The ratio $a : b : c$ becomes:

$$a : \frac{5}{4}a : \frac{3}{2}a$$

Multiplying by $4$ to clean up the fractions, we arrive at the elegant ratio:

**$4 : 5 : 6$**

And there it is. Through the interplay of trigonometry and algebra, we have unraveled the mystery of the triangle. Keep practicing this flow—the ability to switch between geometric rules and algebraic manipulation is the hallmark of a true problem solver.

Animated Solution for Mathematics - Trigonometry: If the lengths of the sides of a triangle are in A.P. and the greatest angle is double the smallest, then a ratio of lengths of the sides of this triangle is :

Select Answer:

Visualized Solution

Visualizing the Triangle

The A.P. Condition

The Angle Constraint

Applying the Sine Rule

Substituting the Angle

Double Angle Expansion

The Cosine Rule Bridge

Equating the Expressions

Algebraic Simplification

Factoring the Equation

Rejecting the Trivial Case

Bringing Back the A.P. Condition

Simplifying to a Quadratic

Solving the Quadratic

Finding the Final Ratio

The Sigma Insight: Properties of Triangles

The Geometry of Harmony

Phase 1

Phase 2

Phase 3

Phase 4

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