一个初三数学问题

显示全部楼层 · 2013-4-7 16:01:07

P点向下交海面于C点，PC=150BC = PB cos65 = PA cos38 = PC cot65 = PC / tan65 = 69.946PB = BC/ cos65 = BC/(sin65/tan65) = 165.508PA = BC/ cos38 = BC/(sin38/tan38) = 88.759AB =PB sin65 - PA sin 38 = 95.4...

千问 · 2013-4-7 16:01:07

解：过P点作PC垂直BC交BA的延长线于C点，过P点作PD垂直BD交于D点，依题意，角CPA=38°，角CPB=角PBD=65°,PC=BD，PD=BC=150,因为tan角PBD=PD/BD,BD=PD/tan角PBD=150/2.1445=69.9464所以AC=PC*tan角CPA=69.9464*0.7813=54.6491所以AB=BC...

千问 · 2013-4-7 16:01:07

150*tan38/tan65=150-AB150*0.7813/2.1445=150-ABAB=150-(150*0.7813/2.1445)=150[1-(0.7813/2.1445)]=95.4米...

千问 · 2013-4-7 16:01:07

我们延长AB 交PD 与D点。 <PDB=90°则<APD =38°
<BPD =65°由图中定义可知tan65°=BD/PD=150/PD=2.1445则PD=69.94米tan38°=AD/PD=AD/69.94=0.7813 则AD =54.6米则AB=DB-AD=150-54.6=95.4米...

千问 · 2013-4-7 16:01:07

AB=150-(150/tan65°)×tan38°=95.3509....