Question

In a triangle ABC, if $\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$ then $\sin A\sin B+\sin B\sin C+\sin C\sin A$ is equal to

• A. $0$
• B. $\frac{9}{4}$
• C. $1$
• D. none of these

Answer 1

Answer: (B)

$\frac{9}{4}$

$\left|\begin{array}{ccc}a& b& c\\ b& c& a\\ c& a& b\end{array}\right|=0$

$R_1\to R_1+R_2+R_3$

$\left|\begin{array}{ccc}a+b+c& a+b+c& a+b+c\\ b& c& a\\ c& a& b\end{array}\right|=0$

$(a+b+c)\left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ b& c& a\\ c& a& b\end{array}\right|=0$

$C_1\to C_1-C_2,C_2\to C_2-C_3$

$\Rightarrow \left|\begin{array}{ccc}0& 0& 1\\ b-c& c-a& a\\ c-a& a-b& b\end{array}\right|=0$

Expanding along $R_1$

$\Rightarrow (c-a{)}^2=(a-b\left)\right(b-c)$

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

$\to a^2+b^2+c^2=ab+bc+ca\Rightarrow 2(a^2+b^2+c^2)-2(ab+bc+ca)$

$\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0\Rightarrow a=b=c$

Hence triangle is equilateral.

$\therefore \sin A\sin B+\sin B\sin C+\sin C\sin A=3\times \frac{3}{4}=\frac{9}{4}$