Question

If $\Delta$ denotes the area of $\Delta ABC$, then $b^{2}sin2C+c^{2}sin2B$ is equal to?

• A. $\Delta$
• B. $2\Delta$
• C. $3\Delta$
• D. $4\Delta$

Answer 1

The correct option is D $4\Delta$

$b^2\sin 2c+c^2\sin 2b$

= $b^2\times 2\sin C\cos C+c^2\times 2\sin B\cos B[\because \sin 2\theta =2\sin \theta \cos \theta ]$

=$b^2\times \frac{c}{R}\cos C+c^2\times \frac{b}{R}\cos B$ $[\because R=\frac{a}{2\sin A}=\frac{b}{2\sin B}=\frac{c}{2\sin C}]$

= $\frac{b^{2}c}{R}\cos C+\frac{b{c}^2}{R}\cos B$

=$\frac{b^{2}c}{R}\left[\frac{a^2+b^2-c^2}{2ab}\right]+\frac{b{c}^2}{R}\left[\frac{a^2+c^2-b^2}{2ac}\right]\left[ByCosineformula\right]$

=$\frac{1}{R}\times \frac{bc}{2a}[a^2+b^2-c^2+a^2+c^2-b^2]$

=$\frac{1}{R}\times \frac{bc}{2a}\times 2a^2$

=$\frac{abc}{2a}$

Now, $R=\frac{abc}{4△}$

$\therefore \frac{abc}{R}=4△$

$\therefore b^2\sin 2C+c^2\sin 2B=4△$

Hence , $4△$ is the correct answer.