Question

If the area of a $\Delta$ABC be $\lambda$ 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

Answer: (C)

$4\lambda$

Given that, $\Delta =\lambda$
Consider, $t=a^2\sin 2B+b^2\sin 2A$
$\Rightarrow t=4R(a\sin A\sin B\cos B+b\sin B\sin A\cos A)$ ...... [ from Sine rule ]
$\Rightarrow t=4R\sin A\sin B(a\cos B+b\cos A)$
$\Rightarrow t=4Rc\sin A\sin B$ ....... [ from projection rule ]
$\Rightarrow t=4Rc\left(\frac{a}{2R}\right)\left(\frac{b}{2R}\right)=\frac{abc}{R}=4\Delta =4\lambda$ ....... [ from sine rule ]
Ans: C