JEE Mains · Maths · STD 11 - Trigonometrical equations
Given \(\frac{{b + c}}{{11}} = \frac{{c + a}}{{12}} = \frac{{a + b}}{{13}}\) for a \(\Delta ABC\) with usual nation. If \(\frac{{\cos \,A}}{\alpha } = \frac{{\cos \,\beta }}{\beta } = \,\frac{{\cos \,C}}{\gamma }\), then the ordered tried \(\left( {\alpha ,\beta ,\gamma } \right)\) has a value
- A \((7, 19, 25)\)
- B \((3, 4, 5)\)
- C \((5, 12, 13)\)
- D \((19, 7, 25)\)
Answer & Solution
Correct Answer
(A) \((7, 19, 25)\)
Step-by-step Solution
Detailed explanation
\(\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\frac{a+b+c}{18}\) \(\Rightarrow a=7 k, b=6 k, c=5 k\) \(\cos A=\frac{b^{2}+c^{2}-a^{2}}{2 b c}=\frac{1}{5}\) \(\cos B=\frac{19}{25}, \cos C=\frac{5}{7}\) \(\frac{1}{5 \alpha}=\frac{19}{35 \beta}=\frac{5}{7 \gamma}\)…
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