Question

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

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

Answer 1

Answer: (B)

We have $\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0$,
$\Rightarrow (c^2-ba)-a(c-a)+b(b-c)=0$
$\Rightarrow c^2-ba-ac+a^2+b^2-bc=0$
$\Rightarrow a^2+b^2+c^2-ab-bc-ac=0$
$\Rightarrow \frac{1}{2}\{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2\}=0$
which is possible only when
$a-b=0,b-c=0,c-a=0$
$\therefore a=b=c$
Then, $\angle A=\angle B=\angle C=60^{\circ }$
${\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=3{\sin }^{2}60^{\circ }$
$=3(\frac{\sqrt{3}}{2}{)}^2=\frac{9}{4}$