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Question
In ΔABC, prove the following:
`(cos A)/a + (cos B)/b + (cos C)/c = (a^2 + b^2 + c^2)/(2abc)`
Solution 1
LHS = `(cos A)/a + (cos B)/b + (cos C)/c`
`= ((("b"^2 + "c"^2 - "a"^2)/"2bc"))/"a" + ((("c"^2 + "a"^2 - "b"^2)/"2ca"))/"b" + ((("a"^2 + "b"^2 - "c"^2)/"2ab"))/"c"`
`= ("b"^2 + "c"^2 - "a"^2)/"2abc" + ("c"^2 + "a"^2 - "b"^2)/"2abc" + ("a"^2 + "b"^2 - "c"^2)/"2abc"`
`= ("b"^2 + "c"^2 - "a"^2 + "c"^2 + "a"^2 - "b"^2 + "a"^2 + "b"^2 - "c"^2)/"2abc"`
`= ("a"^2 + "b"^2 + "c"^2)/"2abc"`
= RHS
Solution 2
LHS = `(cos A)/a + (cos B)/b + (cos C)/c`
= `(b cos A + a cos B)/(ab) + (cos C)/c`
= `c/(ab) + (cos C)/c` ...(By projection rule)
= `c/(ab) + (a^2 + b^2 - c^2)/(2 abc)` ...(By cosine rule)
= `(2c^2 + a^2 + b^2 - c^2)/(2 abc)`
= `(a^2 + b^2 + c^2)/(2 abc)` = R.H.S.
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