Question

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

• A. $\frac{9}{4}$
• B. $\frac{4}{9}$
• C. $1$
• D. $3\sqrt{3}$

Answer 1

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

Given, in$\Delta ABC$, if $\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0$
$\Rightarrow 1(c^2-ab)-a(c-a)+b(b-c)=0\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0\Rightarrow (a^2+b^2-2ab)+(b^2+c^2-2bc)+(c^2+a^2-2ca)=0\Rightarrow (a-b{)}^2+(b-c{)}^2+(c-a{)}^2=0$
Here, sum of squares of three members can be zero if and only if a = b =c
$\Rightarrow \Delta ABC$ is equilateral $\Rightarrow \angle A=\angle B=\angle C={60}^{\circ }\therefore si{n}^{2}A+si{n}^{2}B+si{n}^{2}C=(si{n}^2{60}^{\circ }+si{n}^2{60}^{\circ }+si{n}^2{60}^{\circ })=3\times {\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{9}{4}$