空间角计算,如图如图所示,求角X
4个回答
设AB=x,BC=y,CD=z
连接AD,易知：AD为体对角线,则AD^2=x^2+y^2+z^2
而AO=AB/cosa=x/cosa,DO=DC/cosb=z/cosb
故cosX=(AO^2+DO^2-AD^2)/2AO*DO=[(x^2/cos^2a +z^2/cos^2b -x^2-y^2-z^2)]/[2x/cosa *z/cosb]
易知：x*tana +z*tanb=y
故x^2tan^2a+z^2tan^2b+2xztanatanb=y^2
故x^2/cos^2a +z^2/cos^2b -x^2-y^2-z^2=x^2/cos^2a +z^2/cos^2b -x^2-z^2-x^2tan^2a+z^2tan^2b+2xztanatanb
=-2xzsinasinb/cosacosb
那么cosX=(-2xzsinasinb/cosacosb)/(2xz/(cosacosb)=-sinasinb
所以X=π-arccos(sinasinb)
有点烦,但并不难.
相关问题
