A2=27o, then C+S>0,C−S>0We have 2sin⁡A2=1+sin⁡A−1−sin⁡A.Take, A2=27∘, A=54∘

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$4 \sin 27^{\circ} =$

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$\sqrt{5 - \sqrt{5}} + \sqrt{3 + \sqrt{5}}$

$\sqrt{5 + \sqrt{5}} - \sqrt{3 - \sqrt{5}}$

$\sqrt{5 + \sqrt{5}} + \sqrt{3 + \sqrt{5}}$

$\sqrt{5 + \sqrt{5}} + \sqrt{3 - \sqrt{5}}$

answer is C.

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Detailed Solution

$\frac{A}{2} = 27^{o}, then C + S > 0, C - S > 0$

We have $2 \sin \frac{A}{2} = \sqrt{1 + \sin A} - \sqrt{1 - \sin A}$ .

Take, $\frac{A}{2} = 27^{\circ}, A = 54^{\circ}$

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