Given (b+c)/11 = (c+a)/12 = (a+b)/13 for a ΔABC with usual notation. If cos A/α = cos B/β = cos C/γ,

Q: Why do we introduce the constant $k$? Can't we just solve it directly?

Introducing $k$ is a standard technique to handle continuous proportions. It allows us to express $a, b,$ and $c$ as multiples of a single variable, which makes the subsequent calculations much cleaner and ensures that the final ratio is independent of the triangle's actual size.

Q: I tried using the Sine Rule, but I got stuck. Why is the Cosine Rule better here?

The Sine Rule relates sides to the *sines* of the angles (e.g., $\frac{a}{\sin A} = 2R$). Since the problem asks for the ratio of the *cosines* of the angles, the Cosine Rule is the direct path. It connects sides to cosines, which is exactly what we need.

Q: Does the value of $k$ affect the final answer?

Not at all! Since we are looking for the ratio $\alpha : \beta : \gamma$, any common factor—including $k^2$—will cancel out when you compare the terms. The geometry of the triangle is determined by the *ratios* of the sides, not their absolute lengths.

Question

Given $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$ for a $\Delta ABC$ with usual notation. If $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$, then the ordered triad $(\alpha, \beta, \gamma)$ has a value:

Options:

(A) $(19, 7, 25)$

(B) $(7, 19, 25)$

(C) $(5, 12, 13)$

(D) $(3, 4, 5)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are going to unravel a beautiful problem that sits at the intersection of algebra and geometry. We are given a triangle $\Delta ABC$ where the sums of its sides are in a specific proportion:

$$\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$$

Our mission is to find the triad $(\alpha, \beta, \gamma)$ that governs the ratio of the cosines of the angles. This might look like a daunting task, but let's break it down step by step.

### The Algebraic Dance

Whenever you encounter a continuous proportion like this, the most powerful tool in your arsenal is the proportionality constant. Let's set:

$$\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k$$

This simple step transforms our complex relationship into a system of three linear equations:

1. $b+c = 11k$
2. $c+a = 12k$
3. $a+b = 13k$

If we add these equations together, each variable $a, b,$ and $c$ appears exactly twice. We get:

$$2(a+b+c) = 11k + 12k + 13k = 36k$$

Dividing by two, we find the perimeter of our triangle: $a+b+c = 18k$. This is our **master equation**.

With this, finding the individual sides becomes a simple subtraction game. To find $a$, we subtract the equation for $(b+c)$ from our master equation:

$$a = (a+b+c) - (b+c) = 18k - 11k = 7k$$

Similarly, we find $b = 18k - 12k = 6k$ and $c = 18k - 13k = 5k$. We have successfully unlocked the side lengths of our triangle in terms of $k$: $7k, 6k,$ and $5k$.

### The Bridge to Trigonometry

Now that we have the sides, we use the **Cosine Rule** to find the cosines of the angles. The rule states that $\cos A = \frac{b^2+c^2-a^2}{2bc}$. Substituting our values:

$$\cos A = \frac{(6k)^2 + (5k)^2 - (7k)^2}{2(6k)(5k)} = \frac{36k^2 + 25k^2 - 49k^2}{60k^2} = \frac{12k^2}{60k^2} = \frac{1}{5}$$

We repeat this process for $\cos B$ using the formula $\frac{a^2+c^2-b^2}{2ac}$:

$$\cos B = \frac{(7k)^2 + (5k)^2 - (6k)^2}{2(7k)(5k)} = \frac{49k^2 + 25k^2 - 36k^2}{70k^2} = \frac{38k^2}{70k^2} = \frac{19}{35}$$

Finally, for $\cos C$, we use $\frac{a^2+b^2-c^2}{2ab}$:

$$\cos C = \frac{(7k)^2 + (6k)^2 - (5k)^2}{2(7k)(6k)} = \frac{49k^2 + 36k^2 - 25k^2}{84k^2} = \frac{60k^2}{84k^2} = \frac{5}{7}$$

### The Final Ratio

The problem asks for the triad $(\alpha, \beta, \gamma)$ such that $\frac{\cos A}{\alpha} = \frac{\cos B}{\beta} = \frac{\cos C}{\gamma}$. This implies that the ratio $\alpha : \beta : \gamma$ is identical to the ratio $\cos A : \cos B : \cos C$.

We have the ratio:

$$\frac{1}{5} : \frac{19}{35} : \frac{5}{7}$$

To simplify, we multiply by the least common multiple of the denominators, which is $35$:

$$\left(\frac{1}{5} 	imes 35ight) : \left(\frac{19}{35} 	imes 35ight) : \left(\frac{5}{7} 	imes 35ight) = 7 : 19 : 25$$

The final ordered triad is **$(7, 19, 25)$**.

Animated Solution for Mathematics - Trigonometry: Given $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ for a $Δ A B C$ with usual notation. If $\frac{c o s A}{α} = \frac{c o s B}{β} = \frac{c o s C}{γ}$ , then the ordered triad $(α, β, γ)$ has a value:

Select Answer:

Visualized Solution

Problem Setup

Introducing the Proportionality Constant $k$

Finding the Perimeter Term

Calculating Side $a$

Calculating Side $b$

Calculating Side $c$

Applying the Cosine Rule for $cos A$

Simplifying $cos A$

Calculating $cos B$

Calculating $cos C$

Finding the Ratio $α : β : γ$

Final Calculation and Result

The Sigma Insight: Properties of Triangles

Analyzing the Setup

The Algebraic Dance

The Bridge to Trigonometry

The Final Ratio

Similar Questions

With usual notation, if in a triangle $A B C$ ; $\frac{b + c}{11} = \frac{c + a}{12} = \frac{a + b}{13}$ , then prove that $\frac{c o s A}{7} = \frac{c o s B}{19} = \frac{c o s C}{25}$ .

In a triangle $A B C$ , $B C = 7$ , $A C = 8$ , $A B = α \in N$ and $cos A = \frac{2}{3}$ . If $49 cos (3 C) + 42 = \frac{m}{n}$ , where $gcd (m, n) = 1$ , then $m + n$ is equal to

If in a $Δ A B C$ $a cos^{2} (\frac{C}{2}) + c cos^{2} (\frac{A}{2}) = \frac{3 b}{2}$ , then the sides $a, b$ and $c$

In a triangle $A B C$ , if $cos A + 2 cos B + cos C = 2$ and the lengths of the sides opposite to the angles $A$ and $C$ are 3 and 7 respectively, then $cos A - cos C$ is equal to

In a triangle $A B C$ , $2 a c sin \frac{1}{2} (A - B + C) =$

With the usual notation, in $Δ A B C$ , if $∠ A + ∠ B = 12 0^{\circ}$ , $a = 3 + 1$ and $b = 3 - 1$ , then the ratio $∠ A : ∠ B$ , is :

If in a triangle $A B C$ , $cos A cos B + sin A sin B sin C = 1$ , Show that $a : b : c = 1 : 1 : 2$ .

There exists a triangle $A B C$ satisfying the conditions

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

Let $A B C$ be a triangle such that $∠ A C B = π /6$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value(s) of $x$ for which $a = x^{2} + x + 1, b = x^{2} - 1$ and $c = 2 x + 1$ is (are)

Select Answer:

Visualized Solution

Problem Setup

Introducing the Proportionality Constant .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Finding the Perimeter Term

Calculating Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Calculating Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b

Calculating Side .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c

Applying the Cosine Rule for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosA

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosA

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosB

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosC

Finding the Ratio .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α:β:γ