Question

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

• A. $3\Delta$
• B. $2\Delta$
• C. $4\Delta$
• D. None of these

Answer 1

The correct option is C $4\Delta$

Area of $△ABC=\Delta$,

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

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

$=\frac{a^{2}b}{R}.\left(\frac{a^2+c^2-b^2}{2ac}\right)+\frac{b^{2}a}{R}.\left(\frac{b^2+c^2-a^2}{2bc}\right)$

$=\frac{ab}{2cR}(a^2+c^2-b^2)+\frac{ab}{2cR}(b^2+c^2-a^2)$

$=\frac{a}{2cR}(b{a}^2+b{c}^2-b^3+b{c}^2-b{a}^2)$

$=\frac{a}{2cR}\left(2b{c}^2\right)=\frac{abc}{R}=4\Delta$