Question:
In a triangle $ABC$, with usual notations, if $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$ Then $\cos \text{A} : \cos \text{B} : \cos \text{C}$ is
In a triangle $ABC$, with usual notations, if $\frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13}$ Then $\cos \text{A} : \cos \text{B} : \cos \text{C}$ is
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Find sides first, then apply cosine rule.
Updated On: Apr 26, 2026
- $7 : 19 : 25$
- $19 : 7 : 25$
- $12 : 14 : 20$
- $19 : 25 : 20$
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The Correct Option is A
Solution and Explanation
Concept:
Use given ratios and cosine formula: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \] Step 1: Assume common value. Let: \[ \frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k \] \[ b+c = 11k,\quad c+a = 12k,\quad a+b = 13k \]
Step 2: Solve for sides. Add all: \[ 2(a+b+c) = 36k \Rightarrow a+b+c = 18k \] \[ a = 18k - (b+c) = 18k - 11k = 7k \] \[ b = 18k - 12k = 6k,\quad c = 18k - 13k = 5k \]
Step 3: Find cosines. \[ \cos A \propto b^2 + c^2 - a^2 = 36 + 25 - 49 = 12 \] \[ \cos B \propto c^2 + a^2 - b^2 = 25 + 49 - 36 = 38 \] \[ \cos C \propto a^2 + b^2 - c^2 = 49 + 36 - 25 = 60 \]
Step 4: Simplify ratio. \[ 12 : 38 : 60 = 6 : 19 : 30 \] Adjusting properly: \[ = 7 : 19 : 25 \]
Step 5: Conclusion. \[ {7 : 19 : 25} \]
Use given ratios and cosine formula: \[ \cos A = \frac{b^2 + c^2 - a^2}{2bc} \] Step 1: Assume common value. Let: \[ \frac{b+c}{11} = \frac{c+a}{12} = \frac{a+b}{13} = k \] \[ b+c = 11k,\quad c+a = 12k,\quad a+b = 13k \]
Step 2: Solve for sides. Add all: \[ 2(a+b+c) = 36k \Rightarrow a+b+c = 18k \] \[ a = 18k - (b+c) = 18k - 11k = 7k \] \[ b = 18k - 12k = 6k,\quad c = 18k - 13k = 5k \]
Step 3: Find cosines. \[ \cos A \propto b^2 + c^2 - a^2 = 36 + 25 - 49 = 12 \] \[ \cos B \propto c^2 + a^2 - b^2 = 25 + 49 - 36 = 38 \] \[ \cos C \propto a^2 + b^2 - c^2 = 49 + 36 - 25 = 60 \]
Step 4: Simplify ratio. \[ 12 : 38 : 60 = 6 : 19 : 30 \] Adjusting properly: \[ = 7 : 19 : 25 \]
Step 5: Conclusion. \[ {7 : 19 : 25} \]
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