设圆内接连结四边形ABCD,AB=1,BC=2,CD=3,AD=4,连结BD.在三角形ABD和BCD中,根据余弦定理,BD^2=4^2+1-2*4*cosA
BD^2=3^+2^2-2*3*2cosC,圆内接四边形对角之和为180度,A和C互补,
cosA=-cosC,cosA=1/5,BD=√77/5,
sinA=√(1-cosA^2)=2√6/5,
根据正弦定理:2R=BD/sinA,
R=(√77/5)/(2√6/5)/2=√462/24
设圆内接连结四边形ABCD,AB=1,BC=2,CD=3,AD=4,连结BD.在三角形ABD和BCD中,根据余弦定理,BD^2=4^2+1-2*4*cosA
BD^2=3^+2^2-2*3*2cosC,圆内接四边形对角之和为180度,A和C互补,
cosA=-cosC,cosA=1/5,BD=√77/5,
sinA=√(1-cosA^2)=2√6/5,
根据正弦定理:2R=BD/sinA,
R=(√77/5)/(2√6/5)/2=√462/24