The sides of a triangle are three consecutive natural numbers and its largest angle is twice the sma

Q: Why can we cancel $\sin \theta$ from both sides of the equation?

In any triangle, the angles must be greater than $0^\circ$ and less than $180^\circ$. Since $\theta$ is an angle of a triangle, $\sin \theta$ cannot be zero. Therefore, it is perfectly valid to divide both sides by $\sin \theta$ without losing any solutions.

Q: Why do we use both the Sine Law and the Cosine Law?

The Sine Law helps us relate the sides to the angles, but it leaves us with an unknown angle $\theta$. The Cosine Law provides a direct relationship between the sides and the cosine of an angle. By using both, we create a system of equations that allows us to eliminate the trigonometric terms entirely and solve for the side length $n$ algebraically.

Q: Why do we reject the negative root when solving the quadratic equation?

The variable $n$ represents the length of a side of a triangle. By definition, a side length must be a positive real number. Since our quadratic equation yielded $n=4$ and $n=-1$, we must reject $-1$ because a triangle cannot have a side of length $-1$.

Question

The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides of the triangle.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a field, tasked with constructing a triangle. This triangle has sides that are three consecutive natural numbers: $n$, $n+1$, and $n+2$.

There is a hidden elegance here, a constraint that forces the triangle into a specific shape. We are told that the largest angle is exactly twice the smallest one.

Let the smallest angle be $	heta$. Consequently, the largest angle must be $2	heta$.

Geometry dictates that the smallest angle is always opposite the shortest side, and the largest angle is opposite the longest side. Thus, the side of length $n$ is opposite $	heta$, and the side of length $n+2$ is opposite $2	heta$.

### The Sine Law Bridge

To connect these sides to our angles, we turn to the Sine Law. It is the perfect bridge between the algebraic world of side lengths and the trigonometric world of angles.

The law states that for any triangle, the ratio of a side to the sine of its opposite angle is constant. We focus on the two sides we know best:

$$\frac{n}{\sin 	heta} = \frac{n+2}{\sin 2	heta}$$

This equation is our first major milestone. To handle the term $\sin 2	heta$, we use the double-angle identity, $\sin 2	heta = 2 \sin 	heta \cos 	heta$.

Substituting this into our equation, we obtain:

$$\frac{n}{\sin 	heta} = \frac{n+2}{2 \sin 	heta \cos 	heta}$$

Since $	heta$ is an angle in a triangle, $\sin 	heta 
eq 0$, allowing us to cancel it from both sides. This leaves us with a simplified expression for the cosine of the smallest angle:

$$\cos 	heta = \frac{n+2}{2n}$$

### The Cosine Law: A Generalization of Pythagoras

We have one expression for $\cos 	heta$, but we need another to solve for $n$. We use the Cosine Law, which states that for any angle $	heta$ in a triangle with sides $a, b, c$ (where $a$ is opposite $	heta$):

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

In our specific triangle, $a=n$, $b=n+1$, and $c=n+2$. Substituting these values into the formula, we get:

$$\cos 	heta = \frac{(n+1)^2 + (n+2)^2 - n^2}{2(n+1)(n+2)}$$

### The Grand Unification

Since both expressions describe the same angle, they must be equal. This leads us to the following algebraic equation:

$$\frac{n+2}{2n} = \frac{(n+1)^2 + (n+2)^2 - n^2}{2(n+1)(n+2)}$$

We cancel the factor of $2$ from both denominators. Expanding the numerator on the right-hand side, we get $(n^2 + 2n + 1) + (n^2 + 4n + 4) - n^2$, which simplifies to $n^2 + 6n + 5$.

Our equation becomes:

$$\frac{n+2}{n} = \frac{n^2 + 6n + 5}{(n+1)(n+2)}$$

Cross-multiplying gives us:

$$(n+2)(n+1)(n+2) = n(n^2 + 6n + 5)$$

Expanding both sides, the $n^3$ terms cancel out, leaving us with the quadratic equation:

$$n^2 - 3n - 4 = 0$$

Factoring this, we get $(n-4)(n+1) = 0$. Since $n$ must be a natural number, we reject $n=-1$ and accept $n=4$.

The sides of our triangle are **4, 5, and 6**. We have successfully navigated the path from geometry to algebra to find the unique solution.

Animated Solution for Mathematics - Trigonometry: The sides of a triangle are three consecutive natural numbers and its largest angle is twice the smallest one. Determine the sides of the triangle.

Visualized Solution

Define the Sides and Angles

Apply the Sine Law

Substitute Values into the Sine Law

Simplify using Double Angle Formula

Apply the Cosine Law

Substitute Sides into the Cosine Law

Equate the Two Expressions for $cos θ$

Simplify the Numerator

Cross-Multiply to Form a Polynomial

Solve the Quadratic Equation

Determine the Final Sides

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Sine Law Bridge

The Cosine Law

The Grand Unification

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