答:△ADB中,AD=BD=10,AB=17
根据余弦定理得:
cosB=(AB^2+BD^2-AD^2)/(2AB*BD)
=(17^2+10^2-10^2)/(2*17*10)
=17/20
所以:sinB=√[1-(cosB)^2]=√111/20
RT△ACB中:
AC=ABsinB
=17*√111/20
=17√111/20
≈8.955
答:△ADB中,AD=BD=10,AB=17
根据余弦定理得:
cosB=(AB^2+BD^2-AD^2)/(2AB*BD)
=(17^2+10^2-10^2)/(2*17*10)
=17/20
所以:sinB=√[1-(cosB)^2]=√111/20
RT△ACB中:
AC=ABsinB
=17*√111/20
=17√111/20
≈8.955