Question

In $\Delta ABC$, if $a,b$ and $A$ are given, then there are two triangles are formed with the third side $c_1$ and $c_2$ such that the sum of their areas is :

• A. $\frac{1}{2}{b}^2\cos 2A$
• B. $\frac{1}{2}{b}^2\sin 2A$
• C. $b^2\sin 2A$
• D. $b^2\cos 2A$

Answer 1

The correct option is B $\frac{1}{2}{b}^2\sin 2A$

Using cosine rule, we have :
$a^2=b^2+c^2-2bc\cos A$
$\Rightarrow c^2-2bc\cos A+b^2-a^2=0$
It is a quadratic in $c$
$\therefore c_1+c_2=2b\cos A\cdots \left(i\right)$
Now,
$A_1+A_2=\frac{1}{2}b{c}_1\sin A+\frac{1}{2}b{c}_2\sin A$
$\Rightarrow A_1+A_2=\frac{1}{2}b\sin A(c_1+c_2)$
$\Rightarrow A_1+A_2=\frac{1}{2}b\sin A\left(2b\cos A\right)$(From$\left(i\right))$
$\Rightarrow A_1+A_2=\frac{1}{2}{b}^2\sin 2A$