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If twice the square of the diameter of a circle is equal to the sum of the squares of the sides of the inscribed triangle ABC, then sin2 A + sin2B + sin2C is equal to

(A)  2

(B)  3

(C)  4

(D)  1

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Best answer

Correct option  (A) 2

2(2R)2 = a2 + b2 + c2

Now, use sinA = a/2R

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