Delta ABC માં , જો left| {,begin{array}{*{20} | vedclass

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Question

(a) Given, in $\Delta ABC$ $\left| {\,\begin{array}{*{20}{c}}1&a&b\1&c&a\1&b&c\end{array}\,} ight| = 0$

==>$a(c^2 -

Vedclass · Accepted Answer

$\frac{9}{4}$

Vedclass · Answer

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

$\Delta ABC$ માં , જો $\left| {\,\begin{array}{*{20}{c}}1&a&b\\1&c&a\\1&b&c\end{array}\,} \right| = 0$, તો ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C = $

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