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Consider 7 triangles that contains inner vertices angle of star.

180*7 = (70+68+78+82+86+88+∠A) + 2[(180-a)+(180-b)+......(180-g)]

180*7 = (70+68+78+82+86+88+∠A) + 2[(180*7)-(a+b+c+d+e+f+g)]

180*7 = (70+68+78+82+86+88+∠A) + 2[(180*7)-(180*5)]

∠A = 180*3 - 472 = 68
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\(∠DBE + ∠CEB = ∠BDC + ∠ECD\)

The sum of all the interior angles of quadrilateral \(ACDF\) and triangle \(GBE\) is \(360° + 180° = 540°.\)
The measure of the angle \(∠A\) is
\(540° – ( 70° + 68° + 78° + 82° + 86° + 88° ) = 68°.\)

Therefore, the correct answer is C.
Answer: C
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