(Cosine Formulae) if `a ,b ,c` are the lengths of the sides opposite respectively to the angles `A ,B ,C` of a triangle `A B C ,` show that `(i)cosA=(b^2+c^2-a^2)/(2b c)` (ii) `cosB=(c^2+a^2-b^2)/(2a c)` (iii) `cosC=(a^2+b^2-c^2)/(2a b)`

Explanation

`In triangle A B C`
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|vec{{AB}}|={c} ;|vec{{BC}}|={a} ;|vec{{AC}}|={b}`……..(i)

And` vec{{AB}}+vec{{BC}}=vec{{AC}}`…….(2)
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Now, vec{{AB}} cdot vec{{AB}}=(vec{{AC}}-vec{{BC}}) cdot(vec{{AC}}-vec{{BC}}) ` [from (ii)]
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Rightarrow vec{A B} cdot vec{A B}=vec{A C} cdot vec{A C}-vec{A C} cdot vec{B C}-vec{B C} cdot vec{A C}+vec{B C} cdot vec{B C} `
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Rightarrow vec{A B} cdot vec{A B}=vec{A C} cdot vec{A C}+vec{B C} cdot vec{B C}-2 vec{B C} cdot vec{A C} [because vec{A C} cdot vec{B C}=vec{B C} cdot vec{A C}] `
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Rightarrow|vec{A B}|^{2}=|vec{A C}|^{2}+|vec{B C}|^{2}-2 cdot|vec{B C}| cdot|vec{A C}| cdot cos C `
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