
高中解三角形的题
1个回答
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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
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
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