Definitions. Let's take several angles (Fig. 37): ASB, BSC, CSD, which, adjacent sequentially to one another, are located in the same plane around the common vertex S.

Let us rotate the plane of the angle ASB around the common side SB so that this plane makes up some dihedral angle with BSC plane. Then, without changing the resulting dihedral angle, we rotate it around the straight line SC so that the BSC plane makes a certain dihedral angle with the CSD plane. Let's continue this sequential rotation around each common side. If the last side SF coincides with the first side SA, then a figure is formed (Fig. 38), which is called polyhedral angle. Angles ASB, BSC,... are called flat angles or edges, their sides SA, SB, ... are called ribs, and the common vertex S- top polyhedral angle.

Each edge is also an edge of a certain dihedral angle; therefore, in a polyhedral angle there are as many dihedral angles and as many plane angles as there are all the edges in it. Smallest number there are three faces in a polyhedral angle; this angle is called triangular. There may be tetrahedral, pentagonal, etc. angles.

A polyhedral angle is denoted either by a single letter S placed at the vertex, or by a series of letters SABCDE, of which the first denotes the vertex, and the others - the edges in the order of their location.

A polyhedral angle is called convex if it is entirely located on one side of the plane of each of its faces, which is extended indefinitely. This is, for example, the angle shown in drawing 38. On the contrary, the angle in drawing 39 cannot be called convex, since it is located on both sides of the ASB edge or the BCC edge.

If we intersect all the faces of a polyhedral angle with a plane, then a polygon is formed in the section ( abcde ). In a convex polyhedral angle, this polygon is also convex.

We will only consider convex polyhedral angles.

Theorem. In a trihedral angle, each plane angle is less than the sum of the other two plane angles.

Let the largest of the plane angles in the trihedral angle SABC (Fig. 40) be the angle ASC.

Let us plot on this angle the angle ASD, equal to the angle ASB, and draw some straight line AC intersecting SD at some point D. Let us plot SB = SD. By connecting B with A and C, we get \(\Delta\)ABC, in which

AD+DC< АВ + ВС.

Triangles ASD and ASB are congruent because they each contain an equal angle between equal sides: therefore AD = AB. Therefore, if in the derived inequality we discard the equal terms AD and AB, we obtain that DC< ВС.

Now we notice that in triangles SCD and SCB, two sides of one are equal to two sides of the other, but the third sides are not equal; in this case, the larger angle lies opposite the larger of these sides; Means,

∠CSD< ∠ CSВ.

By adding the angle ASD to the left side of this inequality, and the angle ASB equal to it to the right, we obtain the inequality that needed to be proved:

∠ASC< ∠ CSB + ∠ ASB.

We have proven that even the largest plane angle is less than the sum of the other two angles. This means the theorem is proven.

Consequence. Subtract from both sides of the last inequality by angle ASB or angle CSB; we get:

∠ASC - ∠ASB< ∠ CSB;

∠ASC - ∠CSB< ∠ ASB.

Considering these inequalities from right to left and taking into account that angle ASC as the largest of three corners greater than the difference of the other two angles, we come to the conclusion that in a trihedral angle, each plane angle is greater than the difference of the other two angles.

Theorem. In a convex polyhedral angle, the sum of all plane angles is less than 4d (360°) .

Let's intersect the faces (Fig. 41) of the convex angle SABCDE with some plane; from this we get a convex cross-section n-gon ABCDE.

Applying the theorem proved earlier to each of the trihedral angles whose vertices are located at points A, B, C, D and E, we pacholym:

∠ABC< ∠ABS + ∠SВC, ∠BCD < ∠BCS + ∠SCD и т. д.

Let us add up all these inequalities term by term. Then on the left side we get the sum of all angles of the polygon ABCDE, which is equal to 2 dn - 4d , and on the right - the sum of the angles of triangles ABS, SBC, etc., except for those angles that lie at the vertex S. Denoting the sum of these last angles with the letter X , we get after addition:

2dn - 4d < 2dn - x .

Since in differences 2 dn - 4d and 2 dn - x the minuends are the same, then for the first difference to be less than the second, it is necessary that the subtrahend 4 d was more than the deductible X ; that means 4 d > X , i.e. X < 4d .

The simplest cases of equality of trihedral angles

Theorems. Trihedral angles are equal if they have:

1) along an equal dihedral angle enclosed between two correspondingly equal and identically spaced plane angles, or

2) along an equal plane angle enclosed between two correspondingly equal and identically spaced dihedral angles.

1) Let S and S 1 be two trihedral angles (Fig. 42), for which ∠ASB = ∠A 1 S 1 B 1, ∠ASC = ∠A 1 S 1 C 1 (and these equal angles identically located) and the dihedral angle AS is equal to the dihedral angle A 1 S 1 .

Let us insert the angle S 1 into the angle S so that their points S 1 and S, straight lines S 1 A 1 and SA and planes A 1 S 1 B 1 and ASB coincide. Then the edge S 1 B 1 will go along SB (due to the equality of the angles A 1 S 1 B 1 and ASB), the plane A 1 S 1 C 1 will go along ASC (by virtue of the equality of dihedral angles) and the edge S 1 C 1 will go along the edge SC (due to the equality of angles A 1 S 1 C 1 and ASC). Thus, the trihedral angles will coincide with all their edges, i.e. they will be equal.

2) The second sign, like the first, is proved by embedding.

Symmetrical polyhedral angles

As is known, vertical angles are equal when we are talking about angles formed by straight lines or planes. Let's see if this statement is true in relation to polyhedral angles.

Let us continue (Fig. 43) all the edges of the angle SABCDE beyond the vertex S, then another polyhedral angle SA 1 B 1 C 1 D 1 E 1 is formed, which can be called vertical relative to the first angle. It is easy to see that both angles have equal flat and dihedral angles, respectively, but both are located in reverse order. Indeed, if we imagine an observer looking from outside a polyhedral angle at its vertex, then the edges SA, SB, SC, SD, SE will seem to him to be located in a counterclockwise direction, whereas, looking at the angle SA 1 B 1 C 1 D 1 E 1, he sees the edges SA 1, SB 1, ..., located in a clockwise direction.

Polyhedral angles with correspondingly equal plane and dihedral angles, but located in the opposite order, generally cannot be combined when nested; that means they are not equal. Such angles are called symmetrical(relative to vertex S). The symmetry of figures in space will be discussed in more detail below.

Other materials

Let us consider three rays a, b, c, emanating from the same point and not lying in the same plane. A trihedral angle (abc) is a figure made up of three flat angles (ab), (bc) and (ac) (Fig. 2). These angles are called the faces of a trihedral angle, and their sides are called edges; the common vertex of flat angles is called vertex of a trihedral angle.The dihedral angles formed by the faces of a trihedral angle are called dihedral angles of a trihedral angle.

The concept of a polyhedral angle is defined similarly (Fig. 3).

Polyhedron

In stereometry, figures in space called bodies are studied. A visual (geometric) body must be imagined as a part of space occupied physical body and limited by the surface.

A polyhedron is a body whose surface consists of finite number flat polygons (Fig. 4). A polyhedron is called convex if it is located on one side of the plane of every plane polygon on its surface. a common part such a plane and the surface of a convex polyhedron is called a face. The faces of a convex polyhedron are flat convex polygons. The sides of the faces are called the edges of the polyhedron, and the vertices are called the vertices of the polyhedron.

Let us explain this using the example of a familiar cube (Fig. 5). There is a cube convex polyhedron. Its surface consists of six squares: ABCD, BEFC, .... These are its faces. The edges of the cube are the sides of these squares: AB, BC, BE,.... The vertices of a cube are the vertices of the squares: A, B, C, D, E, .... The cube has six faces, twelve edges and eight vertices.

For the simplest polyhedra - prisms and pyramids, which will be the main object of our study - we will give definitions that, in essence, do not use the concept of body. They will be defined as geometric figures indicating all the points in space that belong to them. Concept geometric body and its surface in general case will be given later.

Polyhedral angle

part of space limited by one polyhedral cavity conical surface, the direction of which is a flat polygon without self-intersections. The faces of this surface are called the faces of the mosaic, and the top is called the top of the mosaic. M. u. is called regular if all its linear angles and all its dihedral angles are equal. Meroy M. u. is the area limited by the spherical polygon obtained by the intersection of the faces of the polygon, a sphere with a radius equal to one, and with the center at the vertex of M. y. See also Solid angle.

Big Soviet encyclopedia. - M.: Soviet Encyclopedia. 1969-1978 .

See what a “polyhedral angle” is in other dictionaries:

See solid angle... Big encyclopedic Dictionary

See solid angle. * * * POLYHEDAL ANGLE POLYHEDAL ANGLE, see Solid angle (see SOLID ANGLE) ... encyclopedic Dictionary

Part of space limited by one cavity of a polyhedral conic. surface directing to a swarm of flat polygon without self-intersections. The faces of this surface are called. the edges of the M. u., the top of the apex of the M. u. A polyhedral angle is called correct... Mathematical Encyclopedia

See Solid angle... Natural science. encyclopedic Dictionary

polyhedral angle- math. A part of space bounded by several planes passing through one point (vertex of an angle) ... Dictionary of many expressions

MULTIFACETED, multifaceted, multifaceted (book). 1. Having several faces or sides. Multifaceted stone. Polyhedral angle (a part of space limited by several planes intersecting at one point; mat.). 2. transfer... ... Dictionary Ushakova

- (mat.). If we draw straight lines OA and 0B from point O on a given plane, we obtain angle AOB (Fig. 1). Crap. 1. Point 0 called the vertex of the angle, and straight lines OA and 0B as the sides of the angle. Suppose that two angles ΒΟΑ and Β 1 Ο 1 Α 1 are given. Let us impose them so that... ...

- (mat.). If we draw straight lines OA and 0B from point O on a given plane, we obtain angle AOB (Fig. 1). Crap. 1. Point 0 called the vertex of the angle, and straight lines OA and 0B as the sides of the angle. Suppose that two angles ΒΟΑ and Β1Ο1Α1 are given. Let's superimpose them so that the vertices O... Encyclopedic Dictionary F.A. Brockhaus and I.A. Efron

This term has other meanings, see Angle (meanings). Angle ∠ Dimension ° SI units Radian ... Wikipedia

Flat, geometric figure, formed by two rays (sides of the surface) emanating from one point (the vertex of the surface). Every U. having a vertex at the center O of some circle (central U.), defines on the circle an arc AB bounded by... ... Great Soviet Encyclopedia

POLYHEDAL ANGLES

A polyhedral angle is the spatial analogue of a polygon. Recall that a polygon on a plane is a figure formed by a simple closed broken line and the internal region limited by it. We will consider a ray in space to be an analogue of a point on a plane, and a plane angle in space to be an analogue of a segment on a plane. Then the analogue of a simple closed broken line on the plane is a surface formed by a finite set of plane anglesA 1 S.A. 2 , A 2 S.A. 3 , …, A n -1 SA n, A n SA 1 with a common vertexS (Fig. 1), in which neighboring corners do not have common points, except for the points of a common ray, and non-adjacent corners do not have common points, except for the common vertex. The figure formed by the indicated surface and one of the two parts of space limited by it is called polyhedral angle. Common topScalled top polyhedral angle. RaysS.A. 1 , …, SA nare called ribs polyhedral angle, and the plane angles themselvesA 1 S.A. 2 , A 2 S.A. 3 , …, A n -1 SA n, A n SA 1 – edges polyhedral angle. A polyhedral angle is indicated by the lettersS.A. 1 … A n, indicating the vertex and points on its edges. Depending on the number of faces, polyhedral angles are called trihedral, tetrahedral, pentahedral (Fig. 2), etc.

A polyhedral angle is called convex, if it is a convex figure, i.e. together with any two of its points contains also the one connecting them line segment. In Figure 2, the trihedral and tetrahedral angles are convex, but the pentagonal angle is not.
Let's consider some properties of triangles and similar properties of trihedral angles.
Property 1(Triangle inequality). Each side of a triangle is less than the sum of its other two sides.
A similar property for trihedral angles is the following property.
Property 1". Each plane angle of a trihedral angle is less than the sum of its other two plane angles.
Proof. Consider a trihedral angle SABC . Let the largest of its plane angles be the angle A.S.C.. Then the inequalities hold

ASB ASC< ASC + BSC ;BSC ASC< ASC + ASB .

Thus, it remains to prove the inequality ASC< A.S.B.+ BSC.
Let's put it on the edge A.S.C. corner A.S.D., equal A.S.B. , and period B let's choose so that SB = SD(Fig. 3). Then the triangles A.S.B. And A.S.D. equal (on two sides and the angle between them) and, therefore, AB = AD. Let's use the triangle inequality A.C.< AB + BC . Subtracting from both its parts AD = AB, we get the inequality DC< BC. In triangles DSC And BSC one side is common ( S.C.), SD = SB And DC< BC. In this case, against larger side lies a larger angle and, therefore, DSC< BSC . Adding to both sides of this inequality the angle A.S.D. , equal A.S.B., we obtain the required inequality ASC< A.S.B.+ BSC.

Corollary 1.The sum of the plane angles of a trihedral angle is less than 360° .
Proof. Let SABC– a given trihedral angle. Consider a trihedral angle with vertex A, formed by edges ABS, ACS and angle BAC. Due to the proven property, the inequality holds BAC< BAS+ CAS. Similarly, for trihedral angles with vertices B And WITH there are inequalities: ABC< ABS+ CBS, ACB< ACS+ BCS. Adding these inequalities and taking into account that the sum of the angles of a triangle ABC equal to 180° , we get 180 ° < BAS+CAS+ ABS+CBS+BCS+ ACS= 180 ° - ASB+ 180° - BSC+ 180° - A.S.C.. Hence, ASB+BSC+ASC< 360 ° .
Corollary 2.The sum of the plane angles of a convex polyhedral angle is less than 360.
The proof is similar to the previous one.
Corollary 3.The sum of the dihedral angles of a trihedral angle is greater than 180° .
Proof. Let SABC- triangular angle. Let's choose some point P inside it and drop perpendiculars from it PA 1 , P.B. 1 , PC 1 on the edge (Fig. 4).

Flat corners B 1 PC 1 , A 1 PC 1 , A 1 P.B. 1 complement the corresponding dihedral angles with edges SA, SB, SC up to 180° . Therefore, the sum of these dihedral angles is 540° - ( B 1 PC 1 +A 1 PC 1 + A 1 P.B. 1 ). Considering that the sum of the plane angles of a trihedral with a vertex P angle less than 360° , we find that the sum of the dihedral angles of the original trihedral angle is greater than 180° .
Property 2.The bisectors of a triangle intersect at one point.
Property 2". The bisector planes of the dihedral angles of a trihedral angle intersect along one straight line.
The proof is similar to the plane case. Namely, let SABC- triangular angle. Bisectal plane of a dihedral angle S.A. is the GMT of the angle equidistant from its faces A.S.C. And A.S.B.. Similarly, the bisector plane of a dihedral angle S.B. is the GMT of the angle equidistant from its faces B.S.A. And BSC . The line of their intersection SO will be equidistant from all faces of the trihedral angle and, therefore, the bisector plane of the dihedral angle will pass through it S.C. .
Property 3.The perpendicular bisectors to the sides of a triangle intersect at one point.
Property 3".Planes passing through the bisectors of the faces of a trihedral angle and perpendicular to these faces intersect along one straight line.
The proof is similar to the proof of the previous property.
Property 4.The medians of a triangle intersect at one point.
Property 4".The planes passing through the edges of a trihedral angle and the bisectors of opposite faces intersect along one straight line.
Proof. Consider a trihedral angle SABC,SA=SB=SC(Fig. 5). Then the bisectors S.A. 1 , S.B. 1 , S.C. 1 corners BSC, ASC, ASB are the medians of the corresponding triangles. That's why A.A. 1 , BB 1 , CC 1 – medians of a triangle ABC. Let O– the point of their intersection. Straight SO is contained in all three planes under consideration and, therefore, is the line of their intersection.

Property 5.The altitudes of a triangle intersect at one point.
Property 5". Planes passing through the edges of a trihedral angle and perpendicular to opposite faces intersect along one straight line.
Proof. Consider a trihedral angle with vertex S and ribs a, b, c. Let's denote a 1 , b 1 , c 1 – lines of intersection of faces with planes passing through the corresponding edges and perpendicular to these faces (Fig. 6). Let's fix the point C on the edge c and drop perpendiculars from it C.A. 1 And C.B. 1 on straight lines a 1 and b 1 . Let's denote A And B line intersections C.A. 1 and C.B. 1 with straight lines a And b. Then S.A. 1 is a projection A.A. 1 to the brink BSC. Because B.C. perpendicular S.A. 1 , then it is perpendicular and A.A. 1 . Likewise, A.C. perpendicular BB 1 . Thus, A.A. 1 and BB 1 are the altitudes of the triangle ABC. Let O– the point of their intersection. Planes passing through lines a And a 1 , b And b 1 perpendicular to the plane ABC and, therefore, the line of their intersection SO perpendicular ABC. Means, SO perpendicular AB. On the other side, CO perpendicular AB. Therefore, the plane passing through the edge c And SO will be perpendicular to the opposite edge.
Property 6 (sine theorem). In a triangle ABC with the parties a, b, c accordingly, the equalities take place a : sin A = b: sin B=c: sin C.
Property 6". Let a, b, g - flat angles of a trihedral angle, a, b, c– dihedral angles opposite them. Then sin a : sin a= sin b : sin b= sin g : sin c.
Proof. Let SABC- triangular angle. Let's drop from the point C perpendicular CC 1 to the plane A.S.B. and perpendicular C.A. 1 on the edge S.A.(Fig. 7). Then the angle C.A. 1 C 1 will linear angle dihedral angle a. That's why CC 1 = C.A. 1 sin a = S.C. sin b sin a. Similarly it is shown that CC 1 = CB 1 sin b = SC sin a sin b. Consequently, the equality sin b sin a = sin a sin b and, therefore, the equality sin a : sin a= sinb : sin b. In a similar way it is proved that the equality sin b : sin b= sin g : sin c.

Property 7.If in convex quadrilateral If you can inscribe a circle, then the sums of the opposite sides are equal.
Property 7". If a sphere can be inscribed into a convex tetrahedral angle, then the sums of opposite plane angles are equal.

Literature
1. Hadamard J. Elementary geometry. Part II. Stereometry. – M.: Uchpedgiz, 1938.
2. Perepelkin D.I. Well elementary geometry. Part II. Geometry in space. – M.-L.: Gostekhizdat, 1949.
3. Encyclopedia elementary mathematics. Book IV. Geometry. - M.; 1963.
4. Smirnova I.M. In the world of polyhedra. – M.: Education, 1995.

TEXT TRANSCRIPT OF THE LESSON:

In planimetry, one of the objects of study is an angle.

An angle is a geometric figure consisting of a point - the vertex of the angle and two rays emanating from this point.

Two angles, one side of which is common and the other two are a continuation of one another, are called adjacent in planimetry.

A compass can be thought of as a model of a plane angle.

Let us recall the concept of a dihedral angle.

This is a figure formed by a straight line a and two half-planes c common border And, not belonging to the same plane in geometry is called a dihedral angle. Half-planes are the faces of a dihedral angle. Straight line a is an edge of a dihedral angle.

The roof of the house clearly demonstrates the dihedral angle.

But the roof of the house in figure two is made in the form of a figure formed from six flat angles with a common vertex so that the angles are taken at in a certain order and each pair of adjacent angles, including the first and last, has common side. What is this roof shape called?

In geometry, a figure made up of angles

And the angles from which this angle is made are called plane angles. The sides of plane angles are called the edges of a polyhedral angle. Point O is called the vertex of the angle.

Examples of polyhedral angles can be found in the tetrahedron and parallelepiped.

The faces of the tetrahedron DBA, ABC, DBC form the polyhedral angle BADC. More often it is called a trihedral angle.

In a parallelepiped, the faces AA1D1D, ABCD, AA1B1B form the trihedral angle AA1DB.

Well, the roof of the house is made in the shape of a hexagonal angle. It consists of six flat angles.

A number of properties are true for a polyhedral angle. Let us formulate them and prove them. It says here that the statement

First, for any convex polyhedral angle there is a plane intersecting all its edges.

For proof, consider the polyhedral angle OA1A2 A3…An.

By condition, it is convex. An angle is called convex if it lies on one side of the plane of each of its plane angles.

Since, by condition, this angle is convex, then points O, A1, A2, A3, An lie on one side of the plane OA1A2

Let's carry out midline KM of the triangle OA1A2 and select from the edges OA3, OA4, OAn the edge that forms the smallest dihedral angle with the OKM plane. Let this be edge OAi.(оа total)

Let us consider the half-plane α with the boundary CM, dividing the dihedral angle OKMAi into two dihedral angles. All vertices from A to An lie on one side of the plane α, and point O on the other side. Therefore, the plane α intersects all the edges of the polyhedral angle. The statement has been proven.

Convex polyhedral angles have another important property.

The sum of the plane angles of a convex polyhedral angle is less than 360°.

Consider a convex polyhedral angle with a vertex at point O. By virtue of the proven statement, there is a plane that intersects all its edges.

Let us draw such a plane α, let it intersect the edges of the angle at points A1, A2, A3 and so on An.

The plane α from the outer region of the plane angle will cut off the triangle. The sum of the angles is 180°. We obtain that the sum of all plane angles from A1OA2 to AnOA1 is equal to the expression, we transform this expression, we rearrange the terms, we obtain

IN this expression the sums indicated in brackets are the sums of the plane angles of a trihedral angle, and as is known they are greater than the third plane angle.

This inequality can be written for all trihedral angles forming a given polyhedral angle.

Consequently, we obtain the following continuation of the equality

The answer proves that the sum of the plane angles of a convex polyhedral angle is less than 360 degrees.