Question

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

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

Hint:

For an equilateral triangle, all sides are equal and all angles are \(60^\circ\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Expand the determinant to obtain a relation between \(a, b, c\), and use it to identify the type of triangle.

Step 2: Detailed Explanation:
Given determinant: \[ \begin{vmatrix} 1 & a & b \\ 1 & c & a \\ 1 & b & c \end{vmatrix} = 0 \]
Expanding along the first row: \[ = 1(c^2 - ab) - a(c - a) + b(b - c) \] \[ = c^2 - ab - ac + a^2 + b^2 - bc \]
Rearranging: \[ a^2 + b^2 + c^2 - ab - bc - ca = 0 \]
This can be written as: \[ \frac{1}{2}\left[(a-b)^2 + (b-c)^2 + (c-a)^2\right] = 0 \]
Since sum of squares is zero: \[ (a-b)^2 = (b-c)^2 = (c-a)^2 = 0 \Rightarrow a = b = c \]
Thus, the triangle is equilateral, so: \[ A = B = C = 60^\circ \]
Now, \[ \sin^2 A + \sin^2 B + \sin^2 C = 3 \times \sin^2 60^\circ \] \[ = 3 \times \left(\frac{\sqrt{3}}{2}\right)^2 = 3 \times \frac{3}{4} = \frac{9}{4} \]
Step 3: Final Answer:
\[ \boxed{\frac{9}{4}} \]