ac=2 bc=1 cosc=3/4 求ab的值,求sin(2a+c)的值
问题描述:
ac=2 bc=1 cosc=3/4 求ab的值,求sin(2a+c)的值
答
AB^2=AC^2+BC^2-2AB*BC*cosC=1+4-2*1*2*3/4=2 AB=根号2sinC=根号7/4 sinA=a*sinC/c=根号14/8 sin2A=5根号7/16 cos2A=11/16sin(2a+c)=sin2AcosC+cos2AsinC=15根号7/64+11根号7/64=13根号7/32