It is known that the lengths of the side edges and the bottom edges of the triangular prism abc-a1b1c1 are equal, and the projection of A1 on the bottom ABC is the midpoint of the BC edge, then the out of plane straight line AB and CC1 are formed It is known that the lengths of the side edges and the bottom edges of the triangular prism abc-a1b1c1 are equal, and the projection of A1 on the bottom ABC is the midpoint D of the BC edge, then the cosine value of the angle formed by the out of plane line AB and CC1 is: A、(√3)/4 B、(√5)/4 C、(√7)/4 D、3/4 I see from the Internet that many people's answers are d.3/4, and they all think that a1d ⊥ surface ABC, but according to my personal understanding, the projection point D of A1 on the side of BC can only explain a1d ⊥ BC, not a1d ⊥ surface ABC... What's the matter with projection?

The projection of A1 on the bottom surface ABC is d,
That is to say, make the vertical line of the bottom surface ABC through A1 and intersect with the bottom surface ABC at D

In the RT triangle ABC, the angle a = 90 ° and the bisector of angle B at angle c intersect at point O, then the angle BOC is equal to

135 degrees

In triangle ABC, angle c = 100 degrees, if the bisectors of angle A and angle B intersect at point O, then angle BOC = at point O, then angle BOC=

B + a = 80 ° because C = 100 ° and BOC = 180-80 / 2 = 140 because the angular bisectors of a and B add up to o

In the RT triangle ABC, the angle a = 90 ° and the bisector of angle B at angle c intersect at point O, then the angle BOC is equal to

In triangle ABC, angle c = 100 degrees, if the bisectors of angle A and angle B intersect at point O, then angle BOC = at point O, then angle BOC=

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