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Question
In \[∆ ABC, \frac{b + c}{12} = \frac{c + a}{13} = \frac{a + b}{15}\] Prove that \[\frac{\cos A}{2} = \frac{\cos B}{7} = \frac{\cos C}{11}\]
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Solution
\[\text{ Let } \frac{b + c}{12} = \frac{c + a}{13} = \frac{a + b}{15} = k\]
\[ \Rightarrow b + c = 12k, c + a = 13k, a + b = 15k\]
\[ \Rightarrow b + c + c + a + a + b = 12k + 13k + 15k\]
\[ \Rightarrow 2\left( a + b + c \right) = 40k\]
\[ \Rightarrow a + c + b = 20k\]
\[ \Rightarrow a + 12k = 20k \left( \because b + c = 12k \right)\]
\[ \Rightarrow a = 8k\]
\[\text{ Also }, \]
\[ c + a = 13k\]
\[ \Rightarrow c = 13k - a = 13k - 8k = 5k\]
\[\text{ and } \]
\[a + b = 15k\]
\[ \Rightarrow b = 15k - a = 15k - 8k = 7k\]
\[\text{ Now }, \]
\[\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{k^2 \left( 49 + 25 - 64 \right)}{k^2 \left( 2 \times 35 \right)} = \frac{1}{7}\]
\[\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{k^2 \left( 64 + 25 - 49 \right)}{\left( 2 \times 40 \right) k^2} = \frac{1}{2}\]
\[\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{k^2 \left( 64 + 49 - 25 \right)}{\left( 2 \times 56 \right) k^2} = \frac{11}{14}\]
\[ \therefore \cos A: \cos B: \cos C = \frac{1}{7}: \frac{1}{2}: \frac{11}{14} = 2: 7: 11\]
\[ \Rightarrow \frac{\cos A}{2} = \frac{\cos B}{7} = \frac{\cos C}{11}\]
Hence proved.
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