Sum

With usual notations, prove that `(cos "A")/"a" + (cos "B")/"b" + (cos "C")/"c" = ("a"^2 + "b"^2 + "c"^2)/(2"abc")`

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#### Solution

Consider L.H.S. = `(cos "A")/"a" + (cos "B")/"b" + (cos "C")/"c"`

= `1/"a" (cos "A") + 1/"b" (cos "B") + 1/"c" (cos "C")`

= `1/"a" (("b"^2 + "c"^2 - "a"^2)/(2"bc")) + 1/"b" (("a"^2 + "c"^2 - "b"^2)/(2"ac")) + 1/"c" (("a"^2 + "b"^2 - "c"^2)/(2"ab"))` .......[By consine rule]

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

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

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

= R.H.S.

∴ `(cos "A")/"a" + (cos "B")/"b" + (cos "C")/"c" = ("a"^2 + "b"^2 + "c"^2)/(2"abc")`

Concept: Solutions of Triangle

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