A sophomore geometry math problem, a sophomore geometry math problem. Ask for answer: thank you, tha

Solution: The straight lines a and b of different planes are at an angle of 80 degrees, you may wish to make parallel lines of two straight lines of different planes at the same time through the point p, at this time, the angle of the two straight lines is the angle formed by the coplanar straight line, and the original problem becomes the intersection point p of the two coplanar straight lines, and there are only 2 straight lines equal to the angles formed by these two coplanar straight lines, you may wish to set the equal angle as x

First of all, if the line is coplanar with these two coplanar lines, that is, the line is the bisector of the angle formed by the two lines, so that x=40° or 50° is obtained at this time

Obviously, these two corners are special, so they are discussed according to this boundary.

If the straight line is not coplanar with these two straight lines, then as shown in Figure 1, where Pa is any straight line on the same side of the same side with an angle of 80° formed by these two straight lines (not perpendicular to the plane where the two straight lines are located), ab is the perpendicular line of the plane where the two straight lines are located, the perpendicular foot is b, and bc is the perpendicular line perpendicular to one of the straight lines, and the perpendicular foot is c, obviously bpc 40°. ∵pb=ap*cos∠apb,pc=pb*cos∠bpc=ap* cos∠bpc *cos∠apb pc=ap*cos∠pac

cos pac= cos bpc *cos apb, bpc 40°, cos pac cos40°, the range of angles is (0, 2 ],pac)40°, that is, it is inferred that when the angle between two straight lines is 80°, the minimum angle is coplanar, that is, 40°, and in the same way, when the angle between two straight lines is 100°, the minimum angle is also 50° when it is coplanar, 1When x<40°, it is obvious that there is no such straight line at the point p to satisfy the topic, 2When x = 40°, it is clear that only that angular bisector satisfies the topic, 3

When x (40,50), it is clear that there are two satisfies on the same side of the straight line at an angle of 80° between the two lines, i.e., one on the left and one on the right. 4.When x = 50°, then there are two straight lines with an angle of 100° on the side of the angle bisector satisfies, and at the same time there are two satisfied on the same side of the two straight lines with an angle of 80°, so there are 3 in total, 5

When 90>x>50, obviously there are two on the same side at 80° and two on the same side at 100°, so there are four in total. To sum up, the answer is B.

I hope the landlord is satisfied, if you don't understand, you can ask me on hi, thank you.

What is the angle between the straight line that crosses p and which line?

The center o of the circle is (-2, -2).

In addition, PQ cuts the circle O to the point Q, and the PQ is perpendicular to the Oq;

If pq is the smallest and oq = 1, then po is the smallest;

If PO is the smallest, then the point O is closest to the straight line Y=3, and the PO vertical line Y is the closest;

then po=5, oq=radius=1, pq=root24.;

The minimum tangent length is root number 24.

The center of the circle is (-2, -2).

The square of the distance from the center of the circle to the point p = (m+2) +3+2) = m +4m+29 tangent length squared d = (m+2) +3+2) -r d = m +4m+28

m+2)²+24

d²>=26

d>=2√6

So the minimum tangent length is 2 6

Let the distance d from p to the center of the circle o(-2,-2), the radius of the circle be r=1, the tangent length l, and the tangent point q

then d 2 = (m+2) 2 +25

In the right triangle PQO, L2 = D2 - R2 = (M+2) 2 +24

So when m=-2, l2 is a minimum of 24

That is, the minimum value of the tangent length is 2 times the root number 6

Solution: (1) extend B1A1 and take point E1 on the extension line of B1A1 so that E1m is 4cm; Extend BA, take the point E on the extension line of BA, make EA 2cm, connect EE1, and take the midpoint of EE1 as Q. Connect QM.

It is proved that the intersection of the surface of qm is d, m, n and the surface aa1b1b, the center of the plane aa1b1b is o, the connection fo, the intersection of fo and aa1 is denoted as g, and the connection b1g.

Because, the points f, g, b1, m are in the same plane, fg mb1, fg=mb1=2cm

So, FGB1M is a parallelogram.

So, fg gb1

Because, GB1 DN

So, fg gb1 dn

Therefore, the fg surface aa1b1b, and the fg surface dmn, so fg is the intersection of the surface dmn and the surface aa1b1b.

Extend BB1, extend FM to cross the extension cable of BB1 to H, connect NH.

It is proved that NH is the intersection line between the surface DMN and the surface BB1C1C.

h is a point on the FM extension line, and h is a point on the BB1 extension line, so the H plane B1BCC1 and the H plane DMN

Because n-plane dmn, n-plane b1bcc1

So nh is the intersection of the surface DMN and the surface bb1c1c.

2) From the analysis of the problem, it can be concluded that the intersection point of NH and B1C1 is p connection PM and PN

Because, fe1m hb1m

So, b1h = 1cm

Because, NC1P HB1P

So, B1P=4 3cm, C1P=8 3cmBecause B1M=NC1=2cm, PM+PN=(10+52) 3 cm(2).

Let the polyhedron be an nhedron.

Connecting the center of the circle and the vertices divides the polyhedron into n pyramids, all of which are r in height and have an area of s1 and s2 、......, respectivelysn

s=s1+s2+……sn

v more = 1 3 * r * s1 + ......1/3*r*sn=1/3*r*ss=135

4/3πr³=36π

r = 3v more = 135

(1) Solution: When a=1, let the equation y=kx-k, simultaneous x+(y-2) =1, so that the discriminant is equal to zero, k= 7, so x=1 or y= 7(x-1).

The central angle of the circle is 120 degrees. The arc length of a sector with a radius of 3 cm is 2 * 3 * 120 360 = 2 cm

The area of the sector (equal to the side area): 3 2*120 360 = 3 cm2

The base of the cone is a circle (let its radius be r cm) and its circumference is 2 cm2 = 2 r

r =1 cm

The base area of the cone: 1 2 = cm2Surface area of the cone: 3 = 4 cm2The edge length of the cone is equal to the radius of the sector (3 cm), let the height of the cone be h cmh 2 + r 2 = 3 2

h^2 = 8

h = 2√2 cm

Volume of the cone: r 2 * h 3 = ( 1 2 * 2 ) 3 = (2 2) 3 cm 3

It is important that the cone is fan-shaped along its busbar, and that the side area of the rotating body is only to be manipulated. The arc length of this sector is equal to the circumference of the circumference of the bottom surface of the cone, its radius is equal to the length of the busbar of the cone, the surface area of the cone is equal to the side area of the cone plus its base area, and the volume is 1/3 of the product of the base area of the cone and the height of the cone.

Answer: The side area is 3 square centimeters, the bottom area is square centimeters, and the volume is 2 2 3 cubic centimeters.