Question

In a triangle $ABC$ if $\begin{vmatrix}1&a&b\\ 1&c&a\\ 1&b&c\end{vmatrix} = 0$, 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 $\begin{vmatrix}1&a&b\\ 1&c&a\\ 1&b&c\end{vmatrix} = 0$,
$\Rightarrow (c^2 - ba) - a (c - a) + b (b - c) = 0$
$ \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^2C = 3\sin^2 60^{\circ} $
$ = 3 \bigg(\frac{\sqrt{3}}{2} \bigg)^2 = \frac{9}{4}$