(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 `cosA(b^2+c^2-a^2)/(2b c)` (ii) `cosB(c^2+a^2-b^2)/(2a c)` (iii) (i) `cosC(a^2+b^2-c^2)/(2a b)`

Explanation

Here, components of C are c cos A and c sin A is drawn.

Since, `vec(CD)=b-c cosA`
In `DeltaBDS" "a^(2)=(b-cosA)^(2)+(csinA)^(2)`
`impliesa^(2)=b^(2)+c^(2)cos^(2)A-2bc cosA+c^(2)sin^(2)A`
`implies2bc cosA=b^(2)-a^(2)+c^(2)(cos^(2)Asin^(2)A)`
`:.cosA=(b^(2)+c^(2)-a^(2))/(2ab)`