在平面直角坐标系中,已知△ABC的顶点A(0,-2),C(0,2),顶点B在椭圆y^2/12+x^2/8=1上,则(sinA+sinC)/sinB的值是?
问题描述:
在平面直角坐标系中,已知△ABC的顶点A(0,-2),C(0,2),顶点B在椭圆y^2/12+x^2/8=1上,则(sinA+sinC)/sinB的值是?
答
在平面直角坐标系中,已知△ABC的顶点A(0,-2),C(0,2),顶点B在椭圆y^2/12+x^2/8=1上,则(sinA+sinC)/sinB的值是?
答
椭圆y^2/12+x^2/8=1 焦点坐标A(0,-2),C(0,2),
B在椭圆上
BA/sinC=BC/sinA=AC/sinB=t
BA=t*sinC BC=t*sinA AC=t*sinB
(sinA+sinC)/sinB
=(BC+BA)/AC
=2a/2c
=a/c
=2√3/2
=√3