在△ACD中∠CAD=90°,CD=7,
所以AC=7cosC,BC=BD+CD=2+7=9,
在△ABC中由余弦定理,AB^2=81+49cos^C-126cos^C=81-77cos^C,①
这里,cos^C=(cosC)^2.
3∠B+4∠C=180°,
所以∠B=60°-(4/3)∠C,
∠BAD=∠BAC-90°=90°-(∠B+∠C)=30°+(1/3)∠C,
在△ABD中由正弦定理,2/sin(30°+C/3)=AB/sin(90°-C)=AB/cosC,
所以AB=2cosC/sin(30°+C/3).
代入①,得4cos^C/sin^(30°+C/3)=81-77cos^C,
4(1+cos2C)/[1-cos(60°+2C/3)]=81-(77/2)(1+cos2C),
设x=cos(60°+2C/3),则cos2C=-cos[3(60°+2C/3)]=3x-4x^3,上式变为
4(1+3x-4x^3)/(1-x)=81-(77/2)(1+3x-4x^3),
4(1+4x+4x^2)=81-(77/2)(1+3x-4x^3),②
两边都乘以2,得 8+32x+32x^2=162-77-231x+308x^3,
整理得308x^3-32x^2-263x+77=0,-1/2<x<1/2,
解得x1=11/14(舍),x2=7/22,x3=-1(舍)
由②左,AB=2(1+2x)=36/11.