在海岸A处发现北偏东45度方向

高一数学向量运算题在海岸A处，发现北偏

BC^2=AB^2+AC^2-2AB*ACcos(45+75)° =AB^2+AC^2+AB*AC =[(√3-1)n]^2+(2n)^2+2n*(√3-1)n =6n^2 AC^2=BC^2+AB^2-2BC*ABcosα cosα=(BC^2+AB^2-AC^2)/2BC*AB =√2/2 α=45° 设追及点为D，则∠CBD=180°-45°-45°+30°=120° CD^2=BC^2+BD^2-2BC*BDcos∠CBD (10√3n*t)^2=6n^2+(10n*t)^2-2√6n(10nt)cos120° 300t^2=6+100t^2+10√6t 200t^2-10√6t-...全部

  BC^2=AB^2+AC^2-2AB*ACcos(45+75)° =AB^2+AC^2+AB*AC =[(√3-1)n]^2+(2n)^2+2n*(√3-1)n =6n^2 AC^2=BC^2+AB^2-2BC*ABcosα cosα=(BC^2+AB^2-AC^2)/2BC*AB =√2/2 α=45° 设追及点为D，则∠CBD=180°-45°-45°+30°=120° CD^2=BC^2+BD^2-2BC*BDcos∠CBD (10√3n*t)^2=6n^2+(10n*t)^2-2√6n(10nt)cos120° 300t^2=6+100t^2+10√6t 200t^2-10√6t-6=0 t1=√6/10,t2=-√6/20(舍弃） BD^2=CD^2+BC^2-2CD*BCcosβ (10*√6/10)^2=(10√3*√6/10)^2+6-2(10√3*√6/10)*√6cosβ 6=18+6-12√3cosβ cosβ=√3/2 β=30° 即∠DCB=30° 由α=45°可知BC是一条水平线，可求得缉私船的追及方向为 90°-30°=60°。
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