Question

In a $\Delta ABC$, if $\left| {\begin{array}{*{20}{c}}
1&a&b \\
1&c&a \\
1&b&c
\end{array}} \right| = 0$, then ${\sin ^2}A + {\sin ^2}B + {\sin ^2}C = $
A) $\dfrac{4}{9}$
B) $\dfrac{9}{4}$
C) $\dfrac{{3\sqrt 3 }}{2}$

Answer 1

Hint: First find the value of the determinant. Then multiply the value of determinant by 2 and try to make the formula of ${\left( {a - b} \right)^2},{\left( {b - c} \right)^2},{\left( {c - a} \right)^2}$.
By solving the given equation, then you will get $a = b = c$, which is possible in an equilateral triangle only. After that substitute the value of A, B, and C in the equation and simplify it to get the desired result.

Complete step-by-step answer:
Given: - $\left| {\begin{array}{*{20}{c}}
1&a&b \\
1&c&a \\
1&b&c
\end{array}} \right| = 0$
Take L.H.S. and simplify it.
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&a&b \\
1&c&a \\
1&b&c
\end{array}} \right|$
Subtract row 1 from row 2 and row 3,
${R_2} \to {R_2} - {R_1}$
${R_3} \to {R_3} - {R_1}$
$ \Rightarrow \left| {\begin{array}{*{20}{c}}
1&a&b \\
0&{c - a}&{a - b} \\
0&{b - a}&{c - b}
\end{array}} \right|$
Solve the determinant along the first column (${C_1}$),
$ \Rightarrow 1\left( {\left| {\begin{array}{*{20}{c}}
{c - a}&{a - b} \\
{b - a}&{c - b}
\end{array}} \right|} \right) - 0 + 0$
Solve the determinant inside the bracket,
$ \Rightarrow \left( {c - a} \right)\left( {c - b} \right) - \left( {a - b} \right)\left( {b - a} \right)$
Multiply the terms in the brackets,
$ \Rightarrow \left( {{c^2} - ac - bc + ab} \right) - \left( {ab - {b^2} - {a^2} + ab} \right)$
Open the brackets and change the sign accordingly,
$ \Rightarrow {c^2} - ac - bc + ab - ab + {b^2} + {a^2} - ab$
Now, equate it with R.H.S.,
$ \Rightarrow {a^2} + {b^2} + {c^2} - ab - bc - ac = 0$
Multiply the terms by 2,
$ \Rightarrow 2{a^2} + 2{b^2} + 2{c^2} - 2ab - 2bc - 2ca = 0$
Solving it further, we get the equation as,
$ \Rightarrow {a^2} + {a^2} + {b^2} + {b^2} + {c^2} + {c^2} - 2ab - 2bc - 2ca = 0$
Group the terms,
$ \Rightarrow \left( {{a^2} - 2ab + {b^2}} \right) + \left( {{b^2} - 2bc + {c^2}} \right) + \left( {{c^2} - 2ca +
{a^2}} \right) = 0$ ….. (1)
As we know the formula,
${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$
${\left( {b - c} \right)^2} = {b^2} - 2bc + {c^2}$
${\left( {c - a} \right)^2} = {c^2} - 2ca + {a^2}$
Putting the values of equation (1) in the obtained equation we get the new equation as:
$ \Rightarrow {\left( {a - b} \right)^2} + {\left( {b - c} \right)^2} + {\left( {c - a} \right)^2} = 0$
Since the sum of the square is zero then each term should be zero. Then,
$ \Rightarrow {\left( {a - b} \right)^2} = 0,{\left( {b - c} \right)^2} = 0,{\left( {c - a} \right)^2} = 0$
Take the square root on both sides,
$ \Rightarrow \left( {a - b} \right) = 0,\left( {b - c} \right) = 0,\left( {c - a} \right) = 0$
Move one variable to the other side,
$ \Rightarrow a = b,b = c,c = a$
The above-shown condition is only possible when,
$ \Rightarrow a = b = c$
So, the triangle is equilateral. All angles will be equal,
$ \Rightarrow \angle A = \angle B = \angle C = 60^\circ $
Now, substitute the value of A, B, and C in the equation,
$ \Rightarrow {\sin ^2}A + {\sin ^2}B + {\sin ^2}C = {\sin ^2}60^\circ + {\sin ^2}60^\circ + {\sin
^2}60^\circ $
Add the terms on the right side,
$ \Rightarrow {\sin ^2}A + {\sin ^2}B + {\sin ^2}C = 3{\sin ^2}60^\circ $
Substitute the value of $\sin 60^\circ $,
$ \Rightarrow {\sin ^2}A + {\sin ^2}B + {\sin ^2}C = 3{\left( {\dfrac{{\sqrt 3 }}{2}} \right)^2}$
Square the term and multiply,
\[\therefore \Rightarrow {\sin ^2}A + {\sin ^2}B + {\sin ^2}C = \dfrac{9}{4}\]

Hence, option (B) is the correct answer.

Note: Whenever you are stuck with these types of problems you should always think about which identity, we can use so that we can prove what has been asked like here we have to make the given equation such that we can use the formula of ${\left( {a - b} \right)^2}$. Proceeding like this will make your solution correct.