Question

In the ambiguous case, if two triangles are formed with a, b and A, then prove that the sum of areas of these triangles is $\frac{1}{2}{b}^2\sin 2A.$

Answer 1

Sum of the areas of the triangles
=$\frac{1}{2}ab\sin C_1+\frac{1}{2}ab\sin C_2$
=$\frac{1}{2}ab(\sin C_1+\sin C_2)$ ........(1)
Now from part (b), we have
$c_1+c_2=2b\cos A$
or $k(\sin C_1+\sin C_2)=2k\sin B\cos A$
or $\sin C_1+\sin C_2=2\sin B\cos A$
Hence from (1) ,
sum of the areas of the two triangles
= $\frac{1}{2}ab.2\sin B\cos A$
= $b^2(\frac{a\sin B}{b})\cos A=b^2\sin A\cos A$
=$\frac{1}{2}.2\sin A\cos A=\frac{1}{2}{b}^2\sin 2A.$