If the sum of the squares of the sides of a triangle ABC is equal to twice the square of its circum diameter, then $\sin^{2} A + \sin^{2} B + \sin^{2} C =$

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Detailed Solution

Given $a^{2} + b^{2} + c^{2} = 8 R^{2}$

Use the rules $a = 2 R \sin A . b = 2 R \sin B, c = 2 R \sin C$

Hence,

$\begin{array}{rcl} 4 R^{2} \sin^{2} A + 4 R^{2} \sin^{2} B + 4 R^{2} \sin^{2} C & = & 8 R^{2} \\ \sin^{2} A + \sin^{2} B + \sin^{2} C & = & 2 \end{array}$

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