In a $∆$ABC, a,b, A are given and c₁, c₂ are two values of the third side c. The sum of the areas of two triangles with sides a, b, c₁ and a, b, c₂ is

$\mathrm{Since}, \cos \mathrm{A} = \frac{{\mathrm{c}}^2+{\mathrm{b}}^2-{\mathrm{a}}^2}{2\mathrm{bc}}$
$\Rightarrow$${\mathrm{c}}^2-2\mathrm{bc} \cos \mathrm{A} + {\mathrm{b}}^2-{\mathrm{a}}^2=0$
It is given that, ${\mathrm{c}}_1 \mathrm{and} {\mathrm{c}}_2$ are roots of the equation.
$$\begin{gathered} \therefore {\mathrm{c}}_1+{\mathrm{c}}_2 = 2\mathrm{b} \cos \mathrm{A} \mathrm{and} {\mathrm{c}}_1{\mathrm{c}}_2 = {\mathrm{b}}^2-{\mathrm{a}}^2 \\ \Rightarrow \mathrm{k}(\sin {\mathrm{C}}_1 + \sin {\mathrm{C}}_2) = 2\mathrm{k} \sin \mathrm{B} \cos \mathrm{A} \\ \Rightarrow \sin {\mathrm{C}}_1+\sin {\mathrm{C}}_2 = 2 \sin \mathrm{B} \cos \mathrm{A} \end{gathered}$$
Now, let sum of the areas of two triangles is E.
$$\begin{gathered} \therefore \mathrm{E} = \frac{1}{2}{\mathrm{absinC}}_1+\frac{1}{2}{\mathrm{absinC}}_2 \\ \Rightarrow \mathrm{E} = \frac{1}{2}\mathrm{ab}({\mathrm{sinC}}_1+{\mathrm{sinC}}_2) \\ \Rightarrow \mathrm{E} = \frac{1}{2}\mathrm{ab}(2\sin \mathrm{B} \mathrm{cosA}) \\ \Rightarrow \mathrm{E} = \mathrm{ab} \sin \mathrm{B} \cos \mathrm{A} = ( \mathrm{b}.\sin \mathrm{A}).\mathrm{b}.\cos \mathrm{A} \\ \therefore \mathrm{E} = \frac{1}{2}{\mathrm{b}}^2\sin 2\mathrm{A} \end{gathered}$$