P为正方形ABCD内一点,AP=1 PB=2 PC=3 则角APB=

来源:学生作业帮助网 编辑:作业帮 时间:2024/07/09 08:56:03
P为正方形ABCD内一点,AP=1 PB=2 PC=3 则角APB=
xj@_e1v<>AʲR#B&JKzl >!&Oc~d =v+zzN5<{8]|z|@")Puv\P1wHȲ[{ y oafEdY,s1`; 92:376L&cSIn+ܵqL,Gkp@I8.UKCeL _ˋi^_/ܒ԰eߒ2&Iv-&X.9Qs^|=]UnŸQ#@ ."hP%@kvDTTmPD1ht7D5}=lfIFtڬUty݀b.3nburW=pYȣfO/

P为正方形ABCD内一点,AP=1 PB=2 PC=3 则角APB=
P为正方形ABCD内一点,AP=1 PB=2 PC=3 则角APB=

P为正方形ABCD内一点,AP=1 PB=2 PC=3 则角APB=
设AB=x
由余弦定理
x^2 = PA^2 + PB^2 - 2PA*PBcosAPB = 5-4 cos APB
x^2 = PB^2+PC^2 -2PB*PCcosBPC = 13 - 12cosBPC
AC^2 = 2x^2 = PA^2 +PC^2 - 2PA*PCcos(360-APB-BPC) = 10 - 6cos(APB+BPC)

即可得