课堂达标检测1.已知扇形的半径为R,面积为R2,那么这个扇形圆心角的弧度数是()A.1 B.EMBEDEquation.DSMT4 C.2 D.4【解析】选C.由S扇形=EMBEDEquation.DSMT4αR2可得,R2=EMBEDEquation.DSMT4αR2,所以α=2.2.-247°30′化为弧度是()EMBEDEquation.DSMT4【解析】选D.-247°30′=-247.5°,因为180°=π,所以1°=EMBEDEquation.DSMT4,所以-247.5°=-247.5×EMBEDEquation.DSMT4=EMBEDEquation.DSMT4.3.EMBEDEquation.DSMT4化为度,结果为________.【解析】πrad=180°,则1rad=(EMBEDEquation.DSMT4)°,所以EMBEDEquation.DSMT4=150°.答案:150°4.把-570°化为2kπ+α(0≤α