YLL討論網

#ed_op#DIV#ed_cl#按照數學的習慣, #ed_op#BR#ed_cl#北偏西 10 度, 是極座標上的  90 + 10 = 100 度; #ed_op#BR#ed_cl#西偏南 40 度, 是極座標上的 180 + 40 = 220 度.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#220 - 100 = 120 度.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#半小時, 表示敵艦走了 5 浬, 剛好也是我艦預計追到的位置 T. 如圖:#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#IMG alt="image file name: 2kaf3e29751c.png" src="http://yll.loxa.edu.tw/phpBB2/richedit/upload/2kaf3e29751c.png" border=0#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#補兩條虛線, 弄個直角三角形, 讓眼睛比較好算. #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#我艦預計的航程 x 浬, #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = (12 + 5/2)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# + (5*√3/2)#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl# = 229#ed_op#BR#ed_cl#x = √229 浬.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#半小時, 航行 √229 浬; 每小時就要 2*√229 浬.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

西40度南      

2(229)^1/2  
大約30浬

一艄敵艦在點p沿著北10度西方向以每小時10浬的速度跳竄,我軍當
時在p的西40度南12浬處,問我軍需用多大的速度前進才能在半小時
內趕上?