[數學]my thoughts
由 studentallen 於 星期三 十二月 27, 2006 8:35 pm
#ed_op#DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2kacb46d2308.jpg" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2kacb46d2308.jpg" border=0#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#1.由題知 BF=BC#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2.作角PCB=60度 =>三角形PBC為正三角形 => BP=PC=BC 又BF=BC#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# =>BP=BF#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#3.連PF 因為BP=BF => 角BPF=角BFP=80度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#4.在AC上取CK=PC 角PKC=角KPC=80度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#5.連KF 易知三角形FPC全等於FKC (SAS) =>KF=PF 又角PFK=60度 #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# =>三角形 PFK為正三角形 =>PK=KF=PF#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#6.角PKC=角KPC=80度 角PKC=角PAK+角APK =>角APK =角PAK=40度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# =>AK=PK 又PK=KF=PF =>AK=KF#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#7.角FKC=20度=角AFK+角KAF =>角AFK=角KAF=10度 =>角BAG=30度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# =>角BGA=90度#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#8.所求角GFC=180度-角FGC-角FCG=180-90-70=20度 #ed_op#/DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#