Question

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

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

Answer 1

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

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

$\Rightarrow a^2+b^2+c^2-ab-bc-ac=0$

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

Since, all the three terms in the above addition are positive which cannot be summed up to zero.

Therefore, $(a-b)=(b-c)=(c-a)=0$

$\Rightarrow a=b=c$

Thus, $A=B=C={60}^o={\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=3\times \frac{3}{4}=\frac{9}{4}$