Question

In a $\Delta ABC$, if $\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0$, then $si{n}^{2}A+si{n}^{2}B+si{n}^{2}C$ is equal to

• A. $\frac{3\sqrt{3}}{2}$
• B. $\frac{9}{4}$
• C. $\frac{5}{4}$
• D. 2

Answer 1

The correct option is B $\frac{9}{4}$

We have, $\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}1& a& b\\ 0& c-a& a-b\\ 0& b-a& c-b\end{array}\right|=0$
{applying $R_2\to R_2-R_1$ and $R_3\to R_3-R_1$}
$\Rightarrow (c-a)(c-b)+(a-b{)}^2=0$
$\Rightarrow a^2+b^2+c^2-ab-bc-ca=0$
$\Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0$
$\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$
$\Rightarrow a=b=c$
Thus $\Delta$ $ABC$ is an equilateral triangle,
$\Rightarrow A=B=C=\frac{\pi }{3}$
Therefore, ${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=3{\sin }^2\frac{\pi }{3}=\frac{9}{4}$