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