Question

一道数学题：在△ABC中,BC=10,AB=4√3,∠ABC=30度,点P在直线AC上,点P到直线AB的距离为1,求PC

Answer 1

AC²=AB²+BC²-2AB*BCcosB=(4√3)²+10²-2*4√3*10*(√3/2)=28

CE=BC/2=10/2=5

△ADP∽△AEC

AP/AC=DP/CE

AP/2√7=1/5

AP=2√7/5

PC=AC-AP=2√7-(2√7/5)=8√7/5

PC=AC+AP=2√7+(2√7/5)=12√7/5