If the right angle vertex C of RT △ ABC is in plane a, the hypotenuse ab ‖ a, and the distance between AB and a is root 6, the projective points of a and B in a are A1 and B1 respectively, and A1C = 3, B1C = 4, then ∠ a1cb1=

Because A1B1 is the projection of AB, so Aa1 ⊥ plane α, BB1 ⊥ plane α, so Aa1 = BB1 = √ 6
Because ab ‖ plane α, so ab ‖ A1B1, so A1B1 = ab
In the right triangle bb1c, BC & sup2; = BB1 & sup2; + B1C & sup2; = 22
AC & sup2; = 15 is obtained in right triangle aa1c, so AB & sup2; = 37 is obtained in right triangle ABC
So A1B1 & sup2; = 37
In △ a1b1c,
According to the cosine theorem, cos ∠ a1cb1 = (A1C & sup2; + B1C & sup2; - A1B1 squared) / (2a1c * B1C) = - 1 / 2
Therefore, a1cb1 = 120 degree

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