如图,B、C是河岸边亮点,A是对岸边上一点,测得角ABC=45°,角ACB=60°,BC=60米,甲想从A点出发在最短的时?

ac 大角对大边

∠CAB=180-60-45=75度,BC/sin75=AC/sin45,AC=sin45*BC/sin75.而,sin75=sin(45+30)=sin45*cos30+cos45*sin30=(√6+√2)/4.则AC=sin45*BC/sin75=√2/2*60/[(√6+√2)/4]=60(√3-1)m=43.92米.点A到对岸边的最短距离为:c...