相交成60度的两条直线和一个平面ABCD所成的角分别为30度和45度,求这两条直线在平面内的射影所成的角的余弦值

设PE与平面ABCD成45°,PF与平面ABCD成30°.
过P做PG⊥平面ABCD,垂足为G,连接EG,FG
∴∠PEG = 45°
∴∠PFG = 30°
设PG = a,
∵∠PEG = 45°
∴PE = √2a,EG =a
∵∠PFG = 30°
∴PF = 2a,FG = √3a
∵EPF = 60°
∴EF"
= PE"+PF"-2PE×PFcos∠EPF
= 2a" + 4a" - 2×√2a×2a×(1/2)
= (6-2√2)a"
∴cos∠EGF
= (GE"+GF" -EF")/2GE×GF
= [a" + 3a" - (6-2√2)a"]/2a×√3a
=(2√2-2)/2√3
=[(√6)-(√3)]/3