The correct option is A Δ Δ #160;_ 1 +Δ #160;_ 2 +Δ #160;_ 3 =2 R ^ 2 ( cos B cos C sin A +cos C cos A sin B +cos A cos B sin C #160;) ...

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Equal Angles Subtend Equal Sides

Δ 1 +Δ 2 +Δ 3...

Question

$Δ_{1} + Δ_{2} + Δ_{3} =$

Δ

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2 Δ

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\frac{Δ}{2}

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4 Δ

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Solution

The correct option is A $Δ$
$Δ_{1} + Δ_{2} + Δ_{3} = 2 R^{2} (cos B cos C sin A + cos C cos A sin B + cos A cos B sin C) = 2 R^{2} {cos B sin (A + C) + cos C cos A sin B} = 2 R^{2} {cos B sin B + cos C cos A sin B} = R^{2} sin B {2 cos B + cos (A + C) + cos (A - C)} = R^{2} sin B {cos B + cos (A - C) + (cos (A + C) + cos B)} = 2 R^{2} sin A sin B sin C = Δ$

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Δ_{1} = ∣ ∣ ∣ ∣ \begin{matrix} a_{1} & a_{2} & a_{3} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3} \end{matrix} ∣ ∣ ∣ ∣, Δ_{2} = ∣ ∣ ∣ ∣ \begin{matrix} 6 a_{1} & 2 a_{2} & 2 a_{3} 3 b_{1} & b_{2} & b_{3} 12 c_{1} & 4 c_{2} & 4 c_{3} \end{matrix} ∣ ∣ ∣ ∣

Δ_{3} = ∣ ∣ ∣ ∣ \begin{matrix} 3 a_{1} + 2 b_{1} - 4 c_{1} & 3 a_{2} + 2 b_{2} - 4 c_{2} & 3 a_{3} + 2 b_{3} - 4 c_{3} 3 b_{1} & 3 b_{2} & 3 b_{3} 3 c_{1} & 3 c_{2} & 3 c_{3} \end{matrix} ∣ ∣ ∣ ∣

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Δ

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I f x^{a} y^{b} = e^{m}, x^{c} y^{d} = e^{n}, Δ_{1} = ∣ ∣ ∣ \begin{matrix} m & b n & d \end{matrix} ∣ ∣ ∣ Δ_{2} = ∣ ∣ ∣ \begin{matrix} a & m c & n \end{matrix} ∣ ∣ ∣

Δ_{3} = ∣ ∣ ∣ \begin{matrix} a & b c & d \end{matrix} ∣ ∣ ∣,

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