如图,在△ABC中,已知B=π/3,AC=4√3,D为BC上一点若AB=AD,试求△ADC的周长的最

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∠ADC=120°.
AD/sinC=CD/sin∠DAC=AC/sin∠ADC=8
AD=8sinC,CD=8sin∠DAC=8sin(180°-C-ADC)=8sin(60°-C)
周长为4√3+8sinC+8sin(60°-C)
=4√3+8sinC+4√3cosC-4sinC
=4√3+4sinC+4√3cosC
=4√3+8sin(C+60°)
因为0°======以下答案可供参考======
供参考答案1:
如图,在△ABC中,已知B=π/3,AC=4√3,D为BC上一点若AB=AD,试求△ADC的周长的最大值(图2)