直线y=3/4X+4交X轴于A,交y轴于B,求角ABO的正弦
人气:309 ℃ 时间:2020-06-17 14:12:50
解答
y=3/4x+4
y=0,x=-16/3
x=0,y=4
所以A(-16/3,0),B)0,4)
所以OA=16/3,OB=4
则由勾股定理
AB=20/3
所以sinABO=OA/AB=4/5
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