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Sine and Cosine Formulae and Their Applications — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSSine and Cosine Formulae and Their Applications4 Marks
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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