Question

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

Answer 1

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