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JEE Mains · Maths · STD 11 - Trigonometrical equations

एक \(\triangle ABC\) में सामान्य संकेतों के आधार पर दिया है कि \(\frac{ b + c }{11}=\frac{ c + a }{12}=\frac{ a + b }{13}\) है। यदि \(\frac{\cos A }{\alpha}=\frac{\cos B }{\beta}=\frac{\cos C }{\gamma}\) है , तो क्रमित त्रिक \((\alpha, \beta, \gamma)\) का एक मान है 

  1. A \((7, 19, 25)\)
  2. B \((3, 4, 5)\)
  3. C \((5, 12, 13)\)
  4. D \((19, 7, 25)\)
Verified Solution

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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