Question

在三角形ABC中,AB=根号6+根号2,角ACB=30度求AC+BC的最大值

Answer 1

由正弦定理：

AB/sinC=2(√6+√2)=AC/sinB=BC/sinA

AC=2(√6+√2)sinB

BC=2(√6+√2)sinA

AC+BC=2(√6+√2)(sinA+sinB)

=2(√6+√2)*2sin[(A+B)/2]cos[(A-B)/2]

=2(√6+√2)*2sin75 cos[(A-B)/2]

=(√6+√2)^2 cos[(A-B)/2]

当 cos[(A-B)/2]＝1,即A＝B时,

最大值是(√6+√2)^2