In ΔABC,sin4⁡A+sin4⁡B+sin4⁡C=sin2⁡Bsin2⁡C+2sin2⁡Csin2⁡A+2sin2⁡Asin2⁡B Then A =

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In $ΔABC, \sin^{4} A + \sin^{4} B + \sin^{4} C = \sin^{2} {Bsin}^{2} C + 2 \sin^{2} {Csin}^{2} A + 2 \sin^{2} {Asin}^{2} B$ Then A =

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$\frac{π}{6}, \frac{5 π}{6}$

$\frac{π}{3}, \frac{5 π}{6}$

$\frac{π}{3}, \frac{π}{2}$

$\frac{π}{4}, \frac{3 π}{4}$

answer is A.

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Detailed Solution

$\begin{array}{l} F r o m s i n e r u l e \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2 R \\ G i v e n t h a t \sin^{4} A + \sin^{4} B + \sin^{4} C = \sin^{2} {Bsin}^{2} C + 2 \sin^{2} {Csin}^{2} A + 2 \sin^{2} {Asin}^{2} B \end{array}$
$\begin{array}{l} a^{4} + b^{4} + c^{4} = b^{2} c^{2} + 2 c^{2} a^{2} + 2 a^{2} b^{2} \\ a^{4} + b^{4} + c^{4} - 2 a^{2} b^{2} - 2 c^{2} a^{2} + 2 b^{2} c^{2} = 3 b^{2} c^{2} \\ {(b^{2} + c^{2} - a^{2})}^{2} = 3 b^{2} c^{2} \\ (2 bccos A)^{2} = 3 b^{2} c^{2} (∵ from cosine rule) \\ 4 b^{2} c^{2} \cos^{2} A = 3 b^{2} c^{2} \\ \cos^{2} A = \frac{3}{4} \\ \cos A = \pm \frac{\sqrt{3}}{2} \Rightarrow A = \frac{π}{6} and \frac{5 π}{6} \end{array}$