连接BD,设BD=x, ∠BAD=θ,则∠BCD=180-θ
cosθ=(AB²+AD²-BD²)/(2AB*AD)=(17-x²)/8
cos∠BCD=-cosθ=(BC²+CD²-BD²)/(2BC*CD)=(13-x²)/12
联立两式,解得
x²=77/5
∴cos∠BAD=cosθ=(17-x²)/8=1/5
cos∠BCD=-cosθ=-1/5
∴sin∠BAD=sin∠BCD=√[1-(1/5)²]=2√6/5
又由正弦定理
BD/sin∠BAD=2R
∴R=(1/2)(BD/sin∠BAD)=(1/2)[√(77/5)/(2√6/5)]=√2310/24