Question

In ∆ABC, if `(2cos "A")/"a" + (cos "B")/"b" + (2cos"C")/"c" = "a"/"bc" + "b"/"ca"`, then show that the triangle is a right angled

Answer 1

In ΔABC by cosine rule, we get

cos A = `("b"^2 + "c"^2 - "a"^2)/(2"bc")`, cos B = `("a"^2 + "c"^2 - "b"^2)/(2"ac")`, cos C = `("a"^2 + "b"^2 - "c"^2)/(2"ab")` 

`(2cos"A")/"a" + (cos "B")/"b" + (2cos"C")/"c" = "a"/"bc" + "b"/"ca"`  .......[Given]

∴ `(2("b"^2 + "c"- "a"^2))/(2"abc") + ("a"^2 + "c"^2 - "b"^2)/(2"abc") + (2("a"^2 + "b"^2 - "c"^2))/(2"abc") = (2"a"^2 + 2"b"^2)/(2"abc")`

∴ 2b² + 2c² – 2a² + a² + c² – b² + 2a² + 2b² – 2c² = 2a² + 2b²

∴ b² – a² + c² = 0

∴ a² = b² + c²

Hence, ΔABC is a right angled triangle.