<p>Given \(\dfrac{b+c}{11} = \dfrac{c+a}{12} = \dfrac{a+b}{13}\), and using the cosine formula with \(\dfrac{\cos A}{\alpha} = \dfrac{\cos B}{\beta} = \dfrac{\cos C}{\gamma}\), find the value of \(475\alpha = 175\beta = 133\gamma\).</p>
Step-by-Step Solution
Key Concept: From the equal ratios of (b+c), (c+a), (a+b), we can find the sides a, b, c in terms of a common parameter. Then using the cosine rule and the given proportionality condition, we establish relationships between α, β, γ to find the required value.
<p><strong>Step 1: Find sides a, b, c</strong></p><p>Let (b+c)/11 = (c+a)/12 = (a+b)/13 = k</p><p>Then: b+c = 11k, c+a = 12k, a+b = 13k</p><p>Adding all three: 2(a+b+c) = 36k, so a+b+c = 18k</p><p>Therefore:</p><p>• a = (a+b+c) - (b+c) = 18k - 11k = 7k</p><p>• b = (a+b+c) - (c+a) = 18k - 12k = 6k</p><p>• c = (a+b+c) - (a+b) = 18k - 13k = 5k</p><p><strong>Step 2: Apply Cosine Rule</strong></p><p>Using cos A = (b² + c² - a²)/(2bc):</p><p>cos A = (36k² + 25k² - 49k²)/(2·6k·5k) = 12k²/60k² = 1/5</p><p>Similarly, cos B = (a² + c² - b²)/(2ac):</p><p>cos B = (49k² + 25k² - 36k²)/(2·7k·5k) = 38k²/70k² = 19/35</p><p>And cos C = (a² + b² - c²)/(2ab):</p><p>cos C = (49k² + 36k² - 25k²)/(2·7k·6k) = 60k²/84k² = 5/7</p><p><strong>Step 3: Use proportionality condition</strong></p><p>Given: cos A/α = cos B/β = cos C/γ</p><p>This means: α : β : γ = cos A : cos B : cos C = 1/5 : 19/35 : 5/7</p><p>Multiplying by 35 (LCM): α : β : γ = 7 : 19 : 25</p><p>Let α = 7m, β = 19m, γ = 25m for some constant m</p><p><strong>Step 4: Find the required value</strong></p><p>We need to find 475α = 175β = 133γ</p><p>475α = 475(7m) = 3325m</p><p>175β = 175(19m) = 3325m</p><p>133γ = 133(25m) = 3325m</p><p>All three expressions equal 3325m. Since the problem asks for this common value with m = 1:</p><p><strong>∴ Answer: 3325</strong></p>
Correct Answer: 3325