(Projection 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)`a=bcosC+ccosB` (ii) `b=ccosA+acosC` (iii)`c=acosB+bcosA`

Explanation

` VEC(AB) ⋅ VEC(AB) =( VEC(AC) − VEC(BC) )⋅( VEC(AC) − VEC(BC) ) `


`=> VEC(AB) ⋅ VEC(AB) = VEC(AC) ⋅ VEC(AC) − VEC(AC) ⋅ VEC(BC) − VEC(BC) ⋅ VEC(AC) + VEC(BC) ⋅ VEC(BC) `


`=> VEC(AB) ⋅ VEC(AB) = VEC(AC) ⋅ VEC(AC) + VEC(BC) ⋅ VEC(BC) −2 VEC(BC) ⋅ VEC(AC) (:. VEC(AC) ⋅ VEC(BC) = VEC(BC) ⋅ VEC(AC) )`


`=> abs( VEC(AB) )^2= abs( VEC(AC) )^2+ abs( VEC(BC) )^2−2⋅ abs( VEC(BC) )⋅ abs( VEC(AC) )⋅cosC`


`=>c^2=b^2+a^2−2⋅a⋅b⋅cosC `


`=>cosC=2 vec(AB) a^2+b^2−c^2`