Question

In a \(\Delta \mathrm{ABC}\), if \(\begin{vmatrix}1 & a & b \\ 1 & c & a \\ 1 & b & c\end{vmatrix}=0\), then \(\sin ^{2} \mathrm{A}+\sin ^{2} \mathrm{B}+\sin ^{2} \mathrm{C}\) is :

• A. \(\dfrac{9}{4}\)
• B. \(\dfrac{5}{4}\)
• C. 2
• D. \(\dfrac{3 \sqrt{3}}{2}\)

Hint:

Expand the determinant to get \(a^2+b^2+c^2=ab+bc+ca\), which forces a = b = c, an equilateral triangle.

Answer 1

The Correct Option is A

Expand along the first row:
\(1(c\cdot c - a\cdot b) - a(1\cdot c - a\cdot 1) + b(1\cdot b - c\cdot 1) = c^2-ab-ac+a^2+b^2-bc\).
Setting this to 0: \(a^2+b^2+c^2=ab+bc+ca\), i.e. \(\tfrac12[(a-b)^2+(b-c)^2+(c-a)^2]=0\).
So a = b = c and the triangle is equilateral with A = B = C = 60 degrees.
Then \(\sin^2A+\sin^2B+\sin^2C = 3\cdot\left(\dfrac{\sqrt3}{2}\right)^2 = \dfrac94\).
Answer: (1) \(\dfrac{9}{4}\)