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Area of Triangle with Coordinates of Vertices Given

If A, B and C...

Question

If A, B and C are the angles of a triangle and $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 + s i n A & 1 + s i n B & 1 + s i n C s i n A + s i n^{2} A & s i n B + s i n^{2} B & s i n C + s i n^{2} C \end{matrix} ∣ ∣ ∣ ∣ = 0$ , which of the following can be an answer?

B = A

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A^{2} = B C

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C = A

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C = B

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Solution

The correct options are
A $B = A$
B $A^{2} = B C$
C $C = B$
D $C = A$
Let $Δ = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 1 + s i n A & 1 + s i n B & 1 + s i n C s i n A + s i n^{2} A & s i n B + s i n^{2} B & s i n C + s i n^{2} C \end{matrix} ∣ ∣ ∣ ∣$
Applying $C_{2} \to C_{2} - C_{1}$ and $C_{3} \to C_{3} - C_{1}$
$Δ = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 + s i n A & s i n B - s i n A & s i n C - s i n A s i n A + s i n^{2} A & (s i n B - s i n A) (s i n B + s i n A + 1) & (s i n C - s i n A) (s i n C + s i n A + 1) \end{matrix} ∣ ∣ ∣ ∣ ∣$
Expanding along $R_{1}$
$∴ Δ = ∣ ∣ ∣ \begin{matrix} s i n B - s i n A & s i n C - s i n A (s i n B - s i n A) (s i n B + s i n A + 1) & (s i n C - s i n A) (s i n C + s i n A + 1) \end{matrix} ∣ ∣ ∣$
$= (s i n B - s i n A) (s i n C - s i n A) ∣ ∣ ∣ \begin{matrix} 1 & 1 s i n B + s i n A + 1 & s i n C + s i n A + 1 \end{matrix} ∣ ∣ ∣$
$= (s i n B - s i n A) (s i n C - s i n A) (s i n C - s i n B)$
But given $Δ = 0$
$∴ (s i n B - s i n A) (s i n C - s i n A) (s i n C - s i n B) = 0$
$∴ s i n B - s i n A = 0$ or $s i n C - s i n A = 0$ or $s i n C - s i n B = 0$
$\Rightarrow s i n B = s i n A$ or $s i n C = s i n A$ or $s i n C = s i n B$
$\Rightarrow B = A$ or $C = A$ or $C = B$

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