C-08-Q189mediumsingle_mcqIf sinα=32\sin\alpha = \dfrac{\sqrt{3}}{2}sinα=23 and π2<α<π\dfrac{\pi}{2} < \alpha < \pi2π<α<π, then α\alphaα equals —aπ3\dfrac{\pi}{3}3πb2π3\dfrac{2\pi}{3}3