If in a ΔABC, $\begin{vmatrix}1 &a& b\\1 &c& a\end{vmatrix}=0$, then the value of $\sin^2 A + \sin^2 B+ \sin^2 C$, is
Answer & explanation
Correct answer: option 1
We have,
$\begin{vmatrix}1 &a& b\\1 &c& a\end{vmatrix}=0$
$⇒a^2 + b^2 + c^2 -ab-bc-ca = 0$
$⇒\frac{1}{2}\left\{(a - b)^2 + (b −c)^2 + (c-a)^2\right\}=0$
$⇒a-b=0,b-c=0, c-a=0⇒ a=b = c$
⇒ ΔABC is equilateral.
$⇒ A=В=С=π/3$
$∴\sin^2 A + \sin^2 B+ \sin^2 C=\frac{3}{4}+\frac{3}{4}+\frac{3}{4}=\frac{9}{4}$