如图,已知矩形ABCD,OA垂直平面ABCD,OA=1,OD与底面ABCD所成的角为30度,OB与CD所成的角为45度,

在矩形ABCD内,过A作BD的垂线,交BD于H.
因OA垂直于矩形ABCD,则OA⊥BD,而BD⊥AH,所以BD⊥△OAH,故BD⊥OH,从而二面角O-BD-A就是角OHA.
△ODA中,因∠ODA = 30°,则AD = OA x tg∠ODA = 1 x (1/√3) = √3
△OBA中,因∠OBA = 45°,则AB = OA x tg∠OBA = 1 x 1) = 1
直角△ABD中,BD = √AD^2+AB^2 = 2,又 BD x AH = AB x AD,2 x AH = 1 x √3,AH = √3/2
直角△OAH中,∠OAH = arctg(OA/AH) = arctg(2/√3) = 49.1°
答:二面角O-BD-A的大小为49.1°