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Trapezoid ABCD is right angle trapezoid, AB / / DC, ∠ DAB = 90 °, PA ⊥ plane ABCD, and PA = ad = DC = & frac12; ab = 1, M is the midpoint of Pb, find the dihedral angle formed by plane AMC and plane BMC
It can be proved that: Ma = MB (if M is used as Mn ‖ PA, Mn ⊥ AB and N is the midpoint of AB);
AC = BC can be proved by connecting AC, so △ AMC ≌ △ BMC,
If AE is perpendicular to cm and connected to be, then be is perpendicular to cm, ∠ AEB is the dihedral angle of AMC and BMC
The results show that AC = radical 2, am = cm = radical 5 / 2, CoSAMC = 1 / 5, sinamc = 2, radical 6 / 5
AE = am * sinamc = radical 30 / 5, cosaeb = - 2 / 3
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