解:射線?$AB$?繞點(diǎn)?$A$?逆時(shí)針轉(zhuǎn)動(dòng)?$t{秒后},$??$∠BAE = 2t°。$?
情況一:?$AB$?轉(zhuǎn)動(dòng)后與?$CD$?在?$EF{同側(cè)時(shí)},$?
?$∠BAF = 180°-∠BAE = 180°-2t°。$?
要使?$AB//CD,$?則?$∠BAF=∠DCF,$?即?$180 - 2t=60,$?解得?$t = 60。$?
情況二:?$AB$?轉(zhuǎn)動(dòng)后與?$CD$?在?$EF{異側(cè)時(shí)},$?
?$∠BAE=∠DCF,$?即?$2t = 60,$?解得?$t = 30。$?
綜上,?$t $?的值為?$30$?或?$60。$?