Question

If the area of a $△ABC$ is $\lambda$ then $a^2\sin 2B+b^2\sin 2A$ is equal to

• A. $2\lambda$
• B. $\lambda$
• C. $4\lambda$
• D. none of these

Answer 1

The correct option is C $4\lambda$

In $△ABC,$

$\sin A=\frac{BC}{AB}=\frac{a}{c}$ and $\sin B=\frac{AC}{AB}=\frac{b}{c}$

$⟹\frac{\sin A}{\sin B}=\frac{a}{b}$

$⟹b\sin A=a\sin B$ ..... $\left(i\right)$

$\cos A=\frac{b}{c}$ and $\cos B=\frac{a}{c}$

$⟹a\cos B+b\cos A=\frac{a^2}{c}+\frac{b^2}{c}=\frac{a^2+b^2}{c}=\frac{c^2}{c}=c$

$\therefore a\cos B+b\cos A=c$ .... $\left(ii\right)$

Consider,

$a^2\sin 2B+b^2\sin 2A$
$=2a^2\sin B\cos B+2b^2\sin A\cos A$
$=2a^2\sin B\cos B+2ab\sin B\cos A$ ...... From $\left(i\right)$
$=2a\sin B(a\cos B+b\cos A)$
$=2ac\sin B$ ..... From $\left(ii\right)$

$=2ab$ ...... $[\because \sin B=\frac{b}{c}]$
$=4\lambda$