Question

If in ∆ABC, $\frac{a\cos A + b\cos B + c\cos C}{a\sin B + b\sin C + c\sin A} = \frac{a + b + c}{9R}$ then the

triangle ABC is

• A. Isosceles
• B. Equilateral
• C. Right angled
• D. None of these

Answer 1

Answer: (B)

Equilateral

Answer 2

LHS $\frac{\Sigma 2R\sin A\cos A}{\frac{ab}{2R} + \frac{bc}{2R} + \frac{ca}{2R} = \frac{2R^{2}(\Sigma\sin 2A)}{ab + bc + ca}}$

⇒$\frac{2R^{2}.4\sin A\sin B\sin C}{\Sigma ab} = \frac{a + b + c}{9R}$

9abc = (a + b + c) (ab + bc + ca) or a(b – c)² + b(c – a)² +

c(a – b)² = 0 ⇒ a = b = c