Question

Rt△ABO的顶点在原点,OA=6,AB=10,OA与x轴正半轴的夹角为30°,求A、B两点的坐标和△ABO的面积

Answer 1

∵AB>OA,∴OA⊥OB,∴|OB|=√(10²-6²)=8

∵OA与x轴正半轴的夹角为30° ∴OB在2,4象限

∴|Xb|=|OB|cos60º=4,|Yb|=|OB|sin60º=4√3

∴A(3√3,3),(-3√3,-3);B(-4,4√3),(4,-4√3)

∴S△ABO=|OA||OB|/2=24