High and high solid geometry detailed calculation process .

1. (Draw your own picture).

Because ab=3, bc=4, ac=5, the known angle abc = 90 degrees.

So ab vertical bc

And because the side abb1a1 is rectangular.

So ab is perpendicular to bb1

So ab is perpendicular to the face bb1c1c

The connection to the BC1 angle AC1B is the angle between AC1 and the side BB1C1C.

From the meaning of the title. BC1 = 4 times the root number 3 (the side BCC1B1 is diamond-shaped, and B1BC=60 degrees,).

ab=3, so the angle ac1b=arctan, the root number of the quarters, number three.

2. Connect to CB1

Deliver BC1 to O

Ab has been shown to be perpendicular to the surface BB1C1C

So AB is perpendicular to CO

And because the side bcc1b1 is diamond-shaped.

So CO is perpendicular to BC1

So CO is perpendicular to the surface ABC1

So CO is the distance from point C to plane ABC1.

It is given according to the title.

The solution yields co=2

Let's do the math a little more, but there should be no problem)

ab should be the diameter.

acb=90°

PO Bottom ABC

po⊥abpa=pb=6

ao=ob=4

po=2√5

Connect to the OMOM PB

om=1/2pb=3

The magnitude of the angle formed by the radius OC and the bus PB is equal to 60°, i.e. MOC = 60°

Cosine theorem.

cos60°=(om +oc -cm) (2*om*oc) to get cm= 13

Over m as mn ab in n

ab//po

Mn = 5, the angle formed by the heterogeneous straight line MC and Po, i.e., cmn

cos cmn=mn cm= 5 13= 65 13 The angle formed by the heterogeneous straight line MC and PO = arccos 65 13 If you agree with my answer, please click "Accept as satisfactory answer" in the lower left corner, and wish you progress in learning!

What does ab and oc mean when they are the radius of the base circle, and what is the center of the bottom circle?

Build a space Cartesian coordinate system, a universal algorithm!

S side = S up + S down = 29

Restore the round table into a cone, let the bus bar of the small cone above be a, and the bus length of the whole cone is b, then (2:5) = (a:b) a = 2 5 b (you can make the height of the cone, so high, the radius of the bottom surface of the line, the bus bar can form two similar triangles, you can see it by drawing a picture).

S side = 2 5 b-2 2 a) = 29 (according to s = lr) The solution is: a = 58 21 b = 145 21

The busbar of the round table = b-a = 29 7

ps: If you choose to fill in the blanks, you can do this: the side area of the round table = (the circumference of the upper bottom surface + the circumference of the lower bottom surface) The bus bar of the round table is 29 = 2 2 + 2 5 ) r then r=29 7).

Set the bottom area of the round platform 9s

Then the upper bottom area is s

The height of the round table is h, so there is 1 3h (s+9s+s*9s under the root number) = 52, that is, sh = 12

Set the connected cone high H

There are similarities available.

h/h+h）^2=s/9s

So h=1 2 h

Therefore, the volume of the cone of this table is 1 3 *h*s=2

<> take BB'Midpoint q

Connect AQ, PQ

Let the edge length of the cube = 2

P is the midpoint of BC and Q is BB'Midpoint.

pq//b'c

APQ is AP vs. B'c.

Cube edge length = 2

aq=√5ap=√5

pq = 2 cosine theorem.

cos∠apq=(ap²+pq²-aq²)/2ap*pq)=√10/10

apq=arccos√10/10

ap vs. b'C into the angle = arccos 10 10 If you agree that I missed the first number of returns, please click "for the satisfactory answer", I wish you progress in learning!

1. This is simple, the side aa1d1d is a square, ae de, cd plane aa1d1d, then cd ae, so ae plane ecd

2. This is also simple, connecting AC1, CD1, plane AEC and plane AD1C is a plane, so the distance is the distance from C1 to plane AD1C, then you can use the equal volume method, CC1D1 and AD1C are the ground of pyramid C1-CAD1, the height can be calculated by one, and the other is required, not counted.