Question

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

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

Answer 1

Answer: (B)

$4\lambda$

$a^2\sin 2B+b^2\sin 2A=2a^2\sin B.\cos B+2b^2\sin A.\cos A$

$=\frac{a^{2}b}{R}\cos B+\frac{b^{2}a}{R}\cos A(\because \frac{\sin A}{a}=\frac{1}{2R}=\frac{\sin B}{b})$

$=\frac{ab}{R}(a\cos B+b\cos A)=\frac{abc}{R}=2bc\sin A$ $[\because c=acosB+bcosA]$

$=4\left(\frac{1}{2}bc\sin A\right)=4\lambda$