Question

In a triangle $ABC$, if $(\sin A+\sin B+\sin C)(\sin A+\sin B-\sin C)=3\sin A\sin B$ is equal to

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{3}$

$\Rightarrow$ $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

$\Rightarrow$ $(\sin A+\sin B+\sin C)(\sin A+\sin B-\sin C)=3\sin A\sin B$

$\Rightarrow$ $(\frac{a}{2R}+\frac{b}{2R}+\frac{c}{2R})(\frac{a}{2R}+\frac{b}{2R}-\frac{c}{2R})=3\times \frac{a}{2R}\times \frac{b}{2R}$

$\Rightarrow$ $\left(\frac{a+b+c}{2R}\right)\left(\frac{a+b-c}{2R}\right)=\frac{3ab}{4R^2}$

$\Rightarrow$ $(a+b+c)(a+b-c)=3ab$

$\Rightarrow$ $(a+b{)}^2-c^2=3ab$

$\Rightarrow$ $a^2+2ab+b^2-c^2=3ab$

$\Rightarrow$ $a^2+b^2-c^2=ab$

$\Rightarrow$ $\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\Rightarrow$ $\cos C=\frac{ab}{2ab}$

$\Rightarrow$ $\cos C=\frac{1}{2}$

$\Rightarrow$ $C={\cos }^{-1}\left(\frac{1}{2}\right)$

$\Rightarrow$ $C=\frac{\pi }{3}$