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किसी $\Delta ABC$ में, यदि $\left| {\,\begin{array}{*{20}{c}}1&a&b\\1&c&a\\1&b&c\end{array}\,} \right| = 0$, तो ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C = $
$\frac{9}{4}$
$\frac{4}{9}$
$1$
$3\sqrt 3 $
Solution
$\Delta ABC$ में, दिया है $\left| {\,\begin{array}{*{20}{c}}1&a&b\\1&c&a\\1&b&c\end{array}\,} \right| = 0$
==>$a(c^2 – ab)-a(c-a)+b(b-c)=0$
==>${a^2} + {b^2} + {c^2} – ab – bc – ca = 0$
==>$2{a^2} + 2{b^2} + 2{c^2} – 2ab – 2bc – 2ca = 0$
==>$({a^2} + {b^2} – 2ab) + ({b^2} + {c^2} – 2bc)$
${(a – b)^2} + {(b – c)^2} + {(c – a)^2} = 0$
यहाँ तीनों अवयवों के वर्गों का योग शून्य हो सकता है, यदि और केवल यदि $a = b = c$
==>$\Delta ABC$ एक समबाहु त्रिभुज है
==>$\angle A = \angle B = \angle C = {60^o}$
$\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}$.
