70o
60o
55o
42o
Correct answer is B
Perimeter of a sector
= 2r + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21
64 = 2 x 21 + \(\frac{\theta}{360^o}\) x 2 x \(\frac{22}{7}\) x 21
64 = 42 + \(\frac{\theta}{360^o}\) x 44 x 3
64 - 42 = \(\frac{\theta}{360^o}\) x 11 x 3
22 = \(\frac{33\theta}{90}\)
\(\theta = \frac{22 \times 30}{11}\)
= 60o
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