In a triangle, the lengths of the two larger sides are 10 and 9, respectively. If the angles are in

Q: Wait, why do we assume the middle angle is $60^\circ$? Is that always true for A.P. angles?

Yes, absolutely! If three angles $A, B, C$ are in A.P., we can write them as $B-d, B, B+d$. Their sum is $(B-d) + B + (B+d) = 3B$. Since the sum of angles in any triangle is $180^\circ$, we have $3B = 180^\circ$, which forces $B = 60^\circ$. It is a beautiful consequence of the A.P. property!

Q: How do I know which side is opposite to the $60^\circ$ angle? The problem just says 'two larger sides are 10 and 9'.

Great observation! In any triangle, the side lengths are proportional to the sines of their opposite angles. The largest side is opposite the largest angle, and the smallest side is opposite the smallest angle. Since $10$ and $9$ are the two larger sides, the third side $c$ must be the smallest. This means $9$ is the middle side, and it must be opposite the middle angle, which we found to be $60^\circ$.

Q: I got two values for $c$ ($5 + \sqrt{6}$ and $5 - \sqrt{6}$). Are both of them correct?

Yes! Both values are mathematically valid. We just need to ensure they satisfy the triangle inequality and our initial assumption that $c$ is the smallest side (i.e., $c < 9$). Since $\sqrt{6} \approx 2.45$, both $5 + 2.45 = 7.45$ and $5 - 2.45 = 2.55$ are less than $9$. Therefore, both are perfectly acceptable solutions.

Question

In a triangle, the lengths of the two larger sides are 10 and 9, respectively. If the angles are in A.P. Then the length of the third side can be

Options:

(A) $5 - \sqrt{6}$

(B) $3\sqrt{3}$

(C) 5

(D) $5 + \sqrt{6}$

(E) none

Sigma Percentile · Accepted Answer

### The Geometry of Harmony: Unlocking the Triangle

Imagine you are standing in the middle of a vast, open field, tasked with constructing a triangle. You are given two sides, $10$ and $9$, but the third side remains a mystery.

You are also told that the angles of this triangle exist in a perfect, rhythmic balance—an Arithmetic Progression. This isn't just a math problem; it is a study in symmetry and constraint. Let us embark on this journey to uncover the hidden side.

### The Symmetry of Angles

When we hear that three angles are in an Arithmetic Progression, our minds should immediately jump to the most elegant representation: $x-d$, $x$, and $x+d$.

When we sum these angles to satisfy the fundamental law of triangles—that the sum of internal angles is $180^\circ$—the common difference $d$ vanishes into thin air.

We are left with $3x = 180^\circ$, which reveals that the middle angle $x$ must be exactly $60^\circ$. This is our anchor; no matter what the other angles are, the middle one is locked in place at $60^\circ$.

### The Logic of Sides

Now, we must place our sides. We know the sides are $10$ and $9$, and we have an unknown side $c$.

Geometry dictates a strict hierarchy: the largest side faces the largest angle, and the smallest side faces the smallest angle. Since $10$ and $9$ are the two larger sides, the unknown side $c$ must be the smallest.

This implies that the side of length $9$ is the middle side, and it must be the one sitting directly opposite our $60^\circ$ angle. We have successfully mapped our triangle's anatomy.

### The Bridge: The Cosine Rule

To connect these pieces, we need a bridge. The Cosine Rule is the ultimate tool for relating three sides and an included angle.

We write it as:

$$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$$

Here, $B = 60^\circ$, $a = 10$, and $b = 9$. Substituting these values, we get:

$$\cos(60^\circ) = \frac{10^2 + c^2 - 9^2}{2(10)c}$$

Since $\cos(60^\circ) = \frac{1}{2}$, our equation becomes:

$$\frac{1}{2} = \frac{100 + c^2 - 81}{20c}$$

### The Quadratic Dance

Now, the algebra begins to sing. Simplifying the numerator, we get $19 + c^2$.

Our equation is now $\frac{1}{2} = \frac{19 + c^2}{20c}$. Cross-multiplying gives us $20c = 2(19 + c^2)$, which simplifies beautifully to $10c = 19 + c^2$.

Rearranging this into the standard quadratic form, we arrive at:

$$c^2 - 10c + 19 = 0$$

Using the quadratic formula $c = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find:

$$c = \frac{10 \pm \sqrt{100 - 76}}{2} = \frac{10 \pm \sqrt{24}}{2}$$

Since $\sqrt{24} = 2\sqrt{6}$, we simplify this to $c = 5 \pm \sqrt{6}$.

### The Final Verification

Before we conclude, we must check our reality. We assumed $c$ was the smallest side, meaning $c < 9$.

With $\sqrt{6} \approx 2.45$, our two possible values are $7.45$ and $2.55$. Both are indeed less than $9$.

We have found two possible triangles that satisfy these conditions. The final values for the third side are **$c = 5 + \sqrt{6}$** and **$c = 5 - \sqrt{6}$**.

Animated Solution for Mathematics - Trigonometry: In a triangle, the lengths of the two larger sides are 10 and 9, respectively. If the angles are in A.P. Then the length of the third side can be

Select Answer:

Visualized Solution

Visualizing the Triangle

Angles in A.P.

Finding the Middle Angle

Identifying the Middle Side

Applying the Cosine Rule

Substituting Values

Simplifying the Equation

Forming the Quadratic Equation

Solving the Quadratic Equation

Final Values of $c$

Verification and Conclusion

The Sigma Insight: Properties of Triangles

The Geometry of Harmony

The Symmetry of Angles

The Logic of Sides

The Bridge

The Quadratic Dance

The Final Verification

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