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Let O be the center of the circle, then \(\widehat{AOC}=2\widehat{ABC}=2.30^0=60^0\)
The answer is : \(12\pi.\dfrac{60}{360}=2\pi\) (in)
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Let O be the center of the circle, then \(\widehat{AOC}=2\widehat{ABC}=2.30^0=60^0\)
The answer is : \(12\pi.\dfrac{60}{360}=2\pi\) (in)
Selected by MathYouLike