Question

在四边形ABCD中,AC平分角DAB,角ABC=60*AC=7,AD=6.三角形ACD的面积=15根号3\2,求AB的长

Answer 1

S(△ACD)=AC*AD*sin∠CAD/2=15√3/2,

sin∠CAD=5√3/14,

cos∠CAD=√[1-(sin∠CAD)^2]=11/14,

sin∠ACB=sin(180°-∠ABC-∠CAB)

=sin(120°-∠CAB)

=sin120°cos∠CAD-cos120°sin∠CAD

=4√3/7.

AB/sin∠ACB=AC/sin∠ABC,

AB=8.