In a triangle ABC if `|(1,a,b),(1,b,c),(1,c,a)|=0` then the value of `sin^2A+sin^2B+sin^2C=`

Options

A. 1
B. `9//4`
C. `4//9`
D. `3sqrt3`

Answer

To solve the problem, we need to analyze the determinant given in the question:

\[ \left| \begin{array}{ccc} 1 & a & b \\ 1 & b & c \\ 1 & c & a \\ \end{array} \right| = 0 \]

This determinant can be expanded using the formula for the determinant of a 3x3 matrix. The determinant can be expressed as follows:

\[ D = 1 \cdot \left| \begin{array}{cc} b & c \\ c & a \\ \end{array} \right| - a \cdot \left| \begin{array}{cc} 1 & c \\ 1 & a \\ \end{array} \right| + b \cdot \left| \begin{array}{cc} 1 & b \\ 1 & c \\ \end{array} \right| \]

Calculating each of these 2x2 determinants:

1. \(\left| \begin{array}{cc} b & c \\ c & a \\ \end{array} \right| = ba - c^2\)

2. \(\left| \begin{array}{cc} 1 & c \\ 1 & a \\ \end{array} \right| = a - c\)

3. \(\left| \begin{array}{cc} 1 & b \\ 1 & c \\ \end{array} \right| = c - b\)

Substituting these back into the determinant \(D\):

\[ D = 1(ba - c^2) - a(a - c) + b(c - b) \]

Expanding this gives:

\[ D = ba - c^2 - a^2 + ac + bc - b^2 \]

Setting the determinant equal to zero:

\[ ba - c^2 - a^2 + ac + bc - b^2 = 0 \]

Rearranging the equation:

\[ a^2 + b^2 + c^2 - ab - ac - bc = 0 \]

This implies that:

\[ \frac{1}{2}((a-b)^2 + (b-c)^2 + (c-a)^2) = 0 \]

Since the sum of squares is zero, it follows that:

\[ a = b = c \]

Now, since \(a\), \(b\), and \(c\) are the sides of a triangle, we can denote them as \(a = b = c = k\) for some positive \(k\).

Next, we need to find the value of \(\sin^2 A + \sin^2 B + \sin^2 C\). In an equilateral triangle:

\[ A = B = C = 60^\circ \]

Thus, we have:

\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]

Calculating \(\sin^2 A\):

\[ \sin^2 A = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]

Therefore:

\[ \sin^2 A + \sin^2 B + \sin^2 C = 3 \cdot \frac{3}{4} = \frac{9}{4} \]

Thus, the final answer is:

\[ \sin^2 A + \sin^2 B + \sin^2 C = \frac{9}{4} \]