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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
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Solution
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.
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