In any , b + c 12 = c + a 13 = a + b 15 , then prove that 2 = 7 = 11 .

Let b + c 12 = c + a 13 = a + b 15 = (say) b + c=12 ,c + a=13 ,a + b=15 (b + c + c + a + a + b)=12 +13 +15 2(a + b + c)=40 a + b + c=20 b + c=12 and a + b + c=20 =8 c + a=13 and a + b + c=20 =7 a + b=15 and a + b + c=20 =5 = b^2+c^2-a^2 2bc =49 ^2+25 ^2-64 ^2 70 ^2 =1 7 = a^2+c^2-b^2 2ac =64 ^2+25 ^2-49 ^2 80 ^2 =1 2 = a^2+b^2-c^2 2ab =64 ^2+49 ^2-25 ^2 112 ^2 =11 14 : : =1 7 :1 2 :11 14 =2:7:11

In any , b + c 12 = c + a 13 = a + b 15 , then prove that 2 = 7 =…

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Question

In any $\triangle\text{ABC},\frac{\text{b + c}}{12}=\frac{\text{c + a}}{13}=\frac{\text{a + b}}{15},$ then prove that $\frac{\cos\text{A}}{2}=\frac{\cos\text{B}}{7}=\frac{\cos\text{C}}{11}.$

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Answer

Let $\frac{\text{b + c}}{12}=\frac{\text{c + a}}{13}=\frac{\text{a + b}}{15}=\lambda\text{(say)}$
$\text{b + c}=12\lambda,\text{c + a}=13\lambda,\text{a + b}=15\lambda$
$(\text{b + c + c + a + a + b})=12\lambda+13\lambda+15\lambda$
$2(\text{a + b + c})=40\lambda$
$\text{a + b + c}=20\lambda$
$\text{b + c}=12\lambda$ and $\text{a + b + c}=20\lambda\Rightarrow\text{a}=8\lambda$
$\text{c + a}=13\lambda$ and $\text{a + b + c}=20\lambda\Rightarrow\text{b}=7\lambda$
$\text{a + b}=15\lambda$ and $\text{a + b + c}=20\lambda\Rightarrow\text{c}=5\lambda$
$\cos\text{A}=\frac{\text{b}^2+\text{c}^2-\text{a}^2}{2\text{bc}}=\frac{49\lambda^2+25\lambda^2-64\lambda^2}{70\lambda^2}=\frac{1}{7}$
$\cos\text{B}=\frac{\text{a}^2+\text{c}^2-\text{b}^2}{2\text{ac}}=\frac{64\lambda^2+25\lambda^2-49\lambda^2}{80\lambda^2}=\frac{1}{2}$
$\cos\text{C}=\frac{\text{a}^2+\text{b}^2-\text{c}^2}{2\text{ab}}=\frac{64\lambda^2+49\lambda^2-25\lambda^2}{112\lambda^2}=\frac{11}{14}$
$\cos\text{A}:\cos\text{B}:\cos\text{C}=\frac{1}{7}:\frac{1}{2}:\frac{11}{14}=2:7:11$

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