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In a $\Delta ABC,$ if $\left| {\,\begin{array}{*{20}{c}}1&a&b\\1&c&a\\1&b&c\end{array}\,} \right| = 0$, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C = $
$\frac{9}{4}$
$\frac{4}{9}$
$1$
$3\sqrt 3 $
Solution
(a) Given, in $\Delta ABC$ $\left| {\,\begin{array}{*{20}{c}}1&a&b\\1&c&a\\1&b&c\end{array}\,} \right| = 0$
==>$a(c^2 – ab)-a(c-a)+b(b-c)=0$
==> ${a^2} + {b^2} + {c^2} – ab – bc – ca = 0$
==> $2{a^2} + 2{b^2} + 2{c^2} – 2ab – 2bc – 2ca = 0$
==> $({a^2} + {b^2} – 2ab) + ({b^2} + {c^2} – 2bc)$
==> ${(a – b)^2} + {(b – c)^2} + {(c – a)^2} = 0$
Here, sum of squares of three members can be zero if and only if $a = b = c$
==> $\Delta ABC$ is equilateral
==> $\angle A = \angle B = \angle C = {60^o}$
$\therefore \,\,{\sin ^2}A + {\sin ^2}B + {\sin ^2}C = $$({\sin ^2}{60^o} + {\sin ^2}{60^o} + {\sin ^2}{60^o})$
$ = \,\,3 \times {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} = \frac{9}{4}$.
