sinB=(根号下2)/2
∠B =45°
则AC=BC
AC=BC=BD+DC
BD=100
tan60°=AC/DC
DC=AC/tan60°
DC=BC/tan60°
DC=(BD+DC)/tan60°
DC=(100+DC)/tan60°
DC=(100+DC)/根号3
DC=100/(根号3-1)
AC=BC=100+100/(根号3-1)
AC=150+50*根号3
设AC=x,由sinB=根号下2/2得角B=45度,所以BC=x(等腰直角三角形),所以DC=x-100,
由∠ADC=60°,得tan∠ADC=AC/DC=x/(x-100)=根号下3,解得AC=150+50根号下3.