Question

In a triangle $ABC$, with usual notations, if $\left|\begin{array}{ccc}1& a& b\\ 1& c& a\\ 1& b& c\end{array}\right|=0$, then ${\sin }^{2}A+24{\sin }^{2}B+36{\sin }^{2}C$ is equal to

Answer 1

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

$\Rightarrow 1(c^2-ab)-a(c-a)+b(b-c)=0$

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

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

$\Rightarrow \frac{1}{2}\left[\right(a-b{)}^2+(b-c{)}^2+(c-a{)}^2]=0$

$\therefore a=b=c$

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

$\therefore \sin A=\sin B=\sin C$

$\therefore A=B=C$

Hence, $\angle A=\angle B=\angle C={60}^o$

${\sin }^{2}A+24{\sin }^{2}B+36{\sin }^{2}C$.

$=\frac{3}{4}+24\times \frac{3}{4}+36\times \frac{3}{4}$

$=\frac{3+72+108}{4}$

$=\frac{183}{4}$.