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In triangle ABC, prove that c=acosB +bc...

In `triangle ABC`, prove that ` c=acosB +bcosA`.

Text Solution

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RHS `=acosB+bcosA`
`=a((c^2+a^2-b^2)/(2ca))+b((b^2+c^2-a^2)/(2bc))` .........[By cosine rule]
`=(c^2+a^2-b^2)/(2c)+(b^2+c^2-a^2)/(2c)`
`=(c^2+a^2-b^2+b^2+c^2-a^2)/(2c)`
`=(2c^2)/(2c)=c=LHS`.
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