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JEE Mains · Maths · STD 11 - Trigonometrical equations

Let \(a , b\) and \(c\) be the length of sides of a triangle \(ABC\) such that \(\frac{ a + b }{7}=\frac{ b + c }{8}=\frac{ c + a }{9}\). If \(r\) and \(R\) are the radius of incircle and radius of circumcircle of the triangle \(ABC\), respectively, then the value of \(\frac{R}{r}\) is equal to

  1. A 2.5
  2. B 2
  3. C 1.5
  4. D 1
Verified Solution

Answer & Solution

Correct Answer

(A) 2.5

Step-by-step Solution

Detailed explanation

\(\frac{ a + b }{7}=\frac{ b + c }{8}=\frac{ c + a }{9}=\lambda\) \(a + b =7 \lambda, b + c =8 \lambda, a + c =9 \lambda\) \(\Rightarrow a + b + c =12 \lambda\) Now \(a =4 \lambda, b =3 \lambda, c =5 \lambda\) \(\because c ^{2}= b ^{2}+ a ^{2}\) \(\angle C =90^{\circ}\)…
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