Question : $a, b, c$ are the lengths of three sides of a triangle ABC. If $a, b, c$ are related by the relation $a^{2}+b^{2}+c^{2}=ab+bc+ca$, then the value of $\sin^{2}A+\sin^{2}B+\sin^{2}C$ is:
Option 1: $\frac{3}{2}$
Option 2: $\frac{3\sqrt3}{2}$
Option 3: $\frac{3}{4}$
Option 4: $\frac{9}{4}$
Correct Answer: $\frac{9}{4}$
Solution :
Given
$a^{2}+b^{2}+c^{2}=ab+bc+ca$
$⇒2a^{2}+2b^{2}+2c^{2}=2ab+2bc+2ca$
$⇒(a^{2}+b^{2}-2ab)+(b^2+c^{2}-2bc)+(a^2+c^2-2ac)=0$
$⇒(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$
$⇒a=b=c$
Which means the triangle is equilateral. So, all angles are of 60º each
Therefore, $\sin^{2}A+\sin^{2}B+\sin^{2}C=\sin^{2}60º+\sin^{2}60º+\sin^{2}60º = \frac{3}{4}+ \frac{3}{4}+ \frac{3}{4}= \frac{9}{4}$
Hence, the correct answer is $\frac{9}{4}$.
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