In `Delta ABC, a, b and A` are given and `c_(1), c_(2)` are two values of the third side c. Prove that the sum of the area of two triangles with sides a, b, `c_(1) and a, b c_(2) " is " (1)/(2) b^(2) sin 2A`

Explanation

We have,
`cos A = (c^(2) + b^(2) -a^(2))/(2bc)`
`rArr c^(2) - 2 bc cos A + b^(2) -a^(2) =0`
It is given that `c_(1) and c_(2)` are the roots of this equation.
Therfore `c_(1) + c_(2) = 2b cos A and c_(1) c_(2) = b^(2) -a^(2)`
`rArr 2R (sin C_(1) + sin C_(2)) = 4 R sin B cos A`
Now, sum of the area of two triangle
`= (1)/(2) ab sin C_(1) + (1)/(2) ab sin C_(2)`
`= (1)/(2) ab (sin C_(1) + sin C_(2))`
`= (1)/(2) ab (2 sin B cos A)`
`= ab sin B cos A`
`= b(b sin A) cos A " " ( :' a sin B = b sin A)`
`=(1)/(2) b^(2) sin 2A`