如图,AB与半径为1的圆0相切于A点,AB=3,AB与圆0的弦AC的夹角为50°。求
AC:
连结OA,作OD⊥AC于D. 因为AB与圆相切于A点,所以∠OAB=90° 则∠0AC=90°=50°-40°. AC=2AD =2OA·cos∠OAC =2cos40°≈1.54.
△ABC的面积.(精确到0.01)
S△ABC=1/2AB·ACsin∠BAC =