一道高中数学题目,求解,谢谢!已知点A(cos70,sin70)B(cos20,sin20),则直线AB的斜率k= ,倾斜角α= ?希望有过程,谢谢!

.直线的斜率k=(sin20-sin70)/(cos20-cos70)=(cos70-cos20)/(cos20-cos70)=-1
倾斜角α= 135度

斜率等于-1; 倾斜角135度
K= (sin70 - sin20)/(cos70 - cos20)
而 sin(70)=cos(90-70)=cos(20); cos(70)=sin(90-20))=sin(20)
故 斜率 k = (cos20-sin20)/(sin20 - cos20) = -1
倾斜角= arctan(k) = arctan(-1) = 135°