One of the simplest spatial figures is polyhedral angles.
A dihedral angle is a figure formed by two half-planes that have a common straight line, which limits them. The half-planes are called the faces of the angle, and the common straight line is called the edge of the angle. The degrees of a dihedral angle are the measure of the corresponding linear angle.
The linear angle of a dihedral angle is the angle formed by two half-lines along which the plane perpendicular to the edge of the dihedral angle intersects the given dihedral angle. The measure of a dihedral angle does not depend on the choice of a linear angle.
A trihedral angle is a figure consisting of three flat angles.
The faces of a trihedral angle are the flat angles, the edges are the sides of the flat angles, the vertex of a trihedral angle is the common vertex of the flat angles.
The dihedral angles formed by the faces of the trihedral angle are called the dihedral angles of the trihedral angle.
Each flat angle of a trihedral angle is less than the sum of its other two flat angles.
A polyhedron is a body whose surface consists of a finite number of plane polygons.
The face of a polyhedron is the surface of each flat polygon.
The edges of a polyhedron are the sides of the faces, the vertices of the polyhedron are the vertices of the faces.
The dihedral angle at an edge of a polyhedron is determined by its faces in which the given edge lies.
A convex polyhedron is one that lies on one side of the plane of each of the flat polygons on its surface.
Each face of a convex polyhedron is a convex polygon. A plane passing through an interior point of a convex polyhedron intersects it and forms a convex polygon in section.
It is interesting. One of the parts of geometry formed a separate science, which is called topology. It studies the topological properties of figures, that is, those that are preserved during continuous deformations of figures "without breaks and gluings."
The theorem of Euler, the great mathematician, physicist and astronomer, formulates the topological property of polyhedra: for any convex polyhedron, the sum of the number of its vertices and the number of faces, excluding the number of its edges, is equal to the number 2.
The concept of a dihedral angle
To introduce the concept of a dihedral angle, first we recall one of the axioms of stereometry.
Any plane can be divided into two half-planes of the line $a$ lying in this plane. In this case, the points lying in the same half-plane are on the same side of the straight line $a$, and the points lying in different half-planes are on opposite sides of the straight line $a$ (Fig. 1).
Picture 1.
The principle of constructing a dihedral angle is based on this axiom.
Definition 1
The figure is called dihedral angle if it consists of a line and two half-planes of this line that do not belong to the same plane.
In this case, the half-planes of the dihedral angle are called faces, and the straight line separating the half-planes - dihedral edge(Fig. 1).
Figure 2. Dihedral angle
Degree measure of a dihedral angle
Definition 2
We choose an arbitrary point $A$ on the edge. The angle between two lines lying in different half-planes, perpendicular to the edge and intersecting at the point $A$ is called linear angle dihedral angle(Fig. 3).
Figure 3
Obviously, every dihedral angle has an infinite number of linear angles.
Theorem 1
All linear angles of one dihedral angle are equal to each other.
Proof.
Consider two linear angles $AOB$ and $A_1(OB)_1$ (Fig. 4).
Figure 4
Since the rays $OA$ and $(OA)_1$ lie in the same half-plane $\alpha $ and are perpendicular to one straight line, they are codirectional. Since the rays $OB$ and $(OB)_1$ lie in the same half-plane $\beta $ and are perpendicular to one straight line, they are codirectional. Hence
\[\angle AOB=\angle A_1(OB)_1\]
Due to the arbitrariness of the choice of linear angles. All linear angles of one dihedral angle are equal to each other.
The theorem has been proven.
Definition 3
The degree measure of a dihedral angle is the degree measure of a linear angle of a dihedral angle.
Task examples
Example 1
Let us be given two non-perpendicular planes $\alpha $ and $\beta $ which intersect along the line $m$. The point $A$ belongs to the plane $\beta $. $AB$ is the perpendicular to the line $m$. $AC$ is perpendicular to the plane $\alpha $ (point $C$ belongs to $\alpha $). Prove that the angle $ABC$ is a linear angle of the dihedral angle.
Proof.
Let's draw a picture according to the condition of the problem (Fig. 5).
Figure 5
To prove this, we recall the following theorem
Theorem 2: A straight line passing through the base of an inclined one, perpendicular to it, is perpendicular to its projection.
Since $AC$ is a perpendicular to the $\alpha $ plane, then the point $C$ is the projection of the point $A$ onto the $\alpha $ plane. Hence $BC$ is the projection of the oblique $AB$. By Theorem 2, $BC$ is perpendicular to an edge of a dihedral angle.
Then, the angle $ABC$ satisfies all the requirements for defining the linear angle of a dihedral angle.
Example 2
The dihedral angle is $30^\circ$. On one of the faces lies the point $A$, which is at a distance of $4$ cm from the other face. Find the distance from the point $A$ to the edge of the dihedral angle.
Decision.
Let's look at Figure 5.
By assumption, we have $AC=4\ cm$.
A-priory degree measure dihedral angle, we have that the angle $ABC$ is equal to $30^\circ$.
Triangle $ABC$ is a right triangle. By definition of the sine of an acute angle
\[\frac(AC)(AB)=sin(30)^0\] \[\frac(5)(AB)=\frac(1)(2)\] \
\(\blacktriangleright\) A dihedral angle is the angle formed by two half-planes and the straight line \(a\) , which is their common boundary.
\(\blacktriangleright\) To find the angle between the planes \(\xi\) and \(\pi\) , you need to find the linear angle spicy or straight) of the dihedral angle formed by the planes \(\xi\) and \(\pi\) :
Step 1: let \(\xi\cap\pi=a\) (the line of intersection of the planes). In the plane \(\xi\) we mark an arbitrary point \(F\) and draw \(FA\perp a\) ;
Step 2: draw \(FG\perp \pi\) ;
Step 3: according to TTP (\(FG\) - perpendicular, \(FA\) - oblique, \(AG\) - projection) we have: \(AG\perp a\) ;
Step 4: The angle \(\angle FAG\) is called the linear angle of the dihedral angle formed by the planes \(\xi\) and \(\pi\) .
Note that the triangle \(AG\) is a right triangle.
Note also that the plane \(AFG\) constructed in this way is perpendicular to both the planes \(\xi\) and \(\pi\) . Therefore, it can be said in another way: angle between planes\(\xi\) and \(\pi\) is the angle between two intersecting lines \(c\in \xi\) and \(b\in\pi\) , forming a plane perpendicular to \(\xi\ ) , and \(\pi\) .
Task 1 #2875
Task level: More difficult than the exam
Dana quadrangular pyramid, all edges of which are equal, and the base is a square. Find \(6\cos \alpha\) , where \(\alpha\) is the angle between its adjacent side faces.
Let \(SABCD\) be a given pyramid (\(S\) is a vertex) whose edges are equal to \(a\) . Therefore, all side faces are equal equilateral triangles. Find the angle between the faces \(SAD\) and \(SCD\) .
Let's draw \(CH\perp SD\) . As \(\triangle SAD=\triangle SCD\), then \(AH\) will also be a height of \(\triangle SAD\) . Therefore, by definition, \(\angle AHC=\alpha\) is the linear dihedral angle between the faces \(SAD\) and \(SCD\) .
Since the base is a square, then \(AC=a\sqrt2\) . Note also that \(CH=AH\) is the height of an equilateral triangle with side \(a\) , hence \(CH=AH=\frac(\sqrt3)2a\) .
Then by the cosine theorem from \(\triangle AHC\) : \[\cos \alpha=\dfrac(CH^2+AH^2-AC^2)(2CH\cdot AH)=-\dfrac13 \quad\Rightarrow\quad 6\cos\alpha=-2.\]
Answer: -2
Task 2 #2876
Task level: More difficult than the exam
The planes \(\pi_1\) and \(\pi_2\) intersect at an angle whose cosine is equal to \(0,2\) . The planes \(\pi_2\) and \(\pi_3\) intersect at a right angle, and the line of intersection of the planes \(\pi_1\) and \(\pi_2\) is parallel to the line of intersection of the planes \(\pi_2\) and \(\ pi_3\) . Find the sine of the angle between the planes \(\pi_1\) and \(\pi_3\) .
Let the line of intersection of \(\pi_1\) and \(\pi_2\) be the line \(a\) , the line of intersection of \(\pi_2\) and \(\pi_3\) be the line \(b\) , and the line of intersection \(\pi_3\) and \(\pi_1\) are the straight line \(c\) . Since \(a\parallel b\) , then \(c\parallel a\parallel b\) (according to the theorem from the section of the theoretical reference “Geometry in space” \(\rightarrow\) “Introduction to stereometry, parallelism”).
Mark the points \(A\in a, B\in b\) so that \(AB\perp a, AB\perp b\) (this is possible because \(a\parallel b\) ). Note \(C\in c\) so that \(BC\perp c\) , hence \(BC\perp b\) . Then \(AC\perp c\) and \(AC\perp a\) .
Indeed, since \(AB\perp b, BC\perp b\) , then \(b\) is perpendicular to the plane \(ABC\) . Since \(c\parallel a\parallel b\) , then the lines \(a\) and \(c\) are also perpendicular to the plane \(ABC\) , and hence to any line from this plane, in particular, to the line \ (AC\) .
Hence it follows that \(\angle BAC=\angle (\pi_1, \pi_2)\), \(\angle ABC=\angle (\pi_2, \pi_3)=90^\circ\), \(\angle BCA=\angle (\pi_3, \pi_1)\). It turns out that \(\triangle ABC\) is rectangular, which means \[\sin \angle BCA=\cos \angle BAC=0,2.\]
Answer: 0.2
Task 3 #2877
Task level: More difficult than the exam
Given lines \(a, b, c\) intersecting at one point, and the angle between any two of them is equal to \(60^\circ\) . Find \(\cos^(-1)\alpha\) , where \(\alpha\) is the angle between the plane formed by the lines \(a\) and \(c\) and the plane formed by the lines \(b\ ) and \(c\) . Give your answer in degrees.
Let the lines intersect at the point \(O\) . Since the angle between any two of them is equal to \(60^\circ\) , then all three lines cannot lie in the same plane. Let us mark a point \(A\) on the line \(a\) and draw \(AB\perp b\) and \(AC\perp c\) . Then \(\triangle AOB=\triangle AOC\) as rectangular in hypotenuse and acute angle. Hence \(OB=OC\) and \(AB=AC\) .
Let's do \(AH\perp (BOC)\) . Then by the three perpendiculars theorem \(HC\perp c\) , \(HB\perp b\) . Since \(AB=AC\) , then \(\triangle AHB=\triangle AHC\) as rectangular along the hypotenuse and leg. Therefore, \(HB=HC\) . Hence, \(OH\) is the bisector of the angle \(BOC\) (since the point \(H\) is equidistant from the sides of the angle).
Note that in this way we have also constructed the linear angle of the dihedral angle formed by the plane formed by the lines \(a\) and \(c\) and the plane formed by the lines \(b\) and \(c\) . This is the angle \(ACH\) .
Let's find this corner. Since we chose the point \(A\) arbitrarily, then let us choose it so that \(OA=2\) . Then in rectangular \(\triangle AOC\) : \[\sin 60^\circ=\dfrac(AC)(OA) \quad\Rightarrow\quad AC=\sqrt3 \quad\Rightarrow\quad OC=\sqrt(OA^2-AC^2)=1.\ ] Since \(OH\) is a bisector, then \(\angle HOC=30^\circ\) , therefore, in a rectangular \(\triangle HOC\) : \[\mathrm(tg)\,30^\circ=\dfrac(HC)(OC)\quad\Rightarrow\quad HC=\dfrac1(\sqrt3).\] Then from rectangular \(\triangle ACH\) : \[\cos\angle \alpha=\cos\angle ACH=\dfrac(HC)(AC)=\dfrac13 \quad\Rightarrow\quad \cos^(-1)\alpha=3.\]
Answer: 3
Task 4 #2910
Task level: More difficult than the exam
The planes \(\pi_1\) and \(\pi_2\) intersect along the line \(l\) , which contains the points \(M\) and \(N\) . The segments \(MA\) and \(MB\) are perpendicular to the line \(l\) and lie in the planes \(\pi_1\) and \(\pi_2\), respectively, and \(MN = 15\) , \(AN = 39\) , \(BN = 17\) , \(AB = 40\) . Find \(3\cos\alpha\) , where \(\alpha\) is the angle between the planes \(\pi_1\) and \(\pi_2\) .
The triangle \(AMN\) is right-angled, \(AN^2 = AM^2 + MN^2\) , whence \ The triangle \(BMN\) is right-angled, \(BN^2 = BM^2 + MN^2\) , whence \ We write the cosine theorem for the triangle \(AMB\): \ Then \ Since the angle \(\alpha\) between the planes is an acute angle, and \(\angle AMB\) turned out to be obtuse, then \(\cos\alpha=\dfrac5(12)\) . Then \
Answer: 1.25
Task 5 #2911
Task level: More difficult than the exam
\(ABCDA_1B_1C_1D_1\) is a parallelepiped, \(ABCD\) is a square with side \(a\) , point \(M\) is the base of the perpendicular dropped from the point \(A_1\) to the plane \((ABCD)\) , moreover, \(M\) is the intersection point of the diagonals of the square \(ABCD\) . It is known that \(A_1M = \dfrac(\sqrt(3))(2)a\). Find the angle between the planes \((ABCD)\) and \((AA_1B_1B)\) . Give your answer in degrees.
We construct \(MN\) perpendicular to \(AB\) as shown in the figure.
Since \(ABCD\) is a square with side \(a\) and \(MN\perp AB\) and \(BC\perp AB\) , then \(MN\parallel BC\) . Since \(M\) is the intersection point of the diagonals of the square, then \(M\) is the midpoint of \(AC\) , therefore, \(MN\) is middle line and \(MN=\frac12BC=\frac(1)(2)a\).
\(MN\) is the projection of \(A_1N\) onto the plane \((ABCD)\) , and \(MN\) is perpendicular to \(AB\) , then, by the three perpendiculars theorem, \(A_1N\) is perpendicular to \(AB \) and the angle between the planes \((ABCD)\) and \((AA_1B_1B)\) is \(\angle A_1NM\) .
\[\mathrm(tg)\, \angle A_1NM = \dfrac(A_1M)(NM) = \dfrac(\frac(\sqrt(3))(2)a)(\frac(1)(2)a) = \sqrt(3)\qquad\Rightarrow\qquad\angle A_1NM = 60^(\circ)\]
Answer: 60
Task 6 #1854
Task level: More difficult than the exam
In the square \(ABCD\) : \(O\) is the intersection point of the diagonals; \(S\) is not in the plane of the square, \(SO \perp ABC\) . Find the angle between the planes \(ASD\) and \(ABC\) if \(SO = 5\) and \(AB = 10\) .
Right triangles \(\triangle SAO\) and \(\triangle SDO\) are equal in two sides and the angle between them (\(SO \perp ABC\) \(\Rightarrow\) \(\angle SOA = \angle SOD = 90^\circ\); \(AO = DO\) , because \(O\) is the point of intersection of the diagonals of the square, \(SO\) is the common side) \(\Rightarrow\) \(AS = SD\) \(\Rightarrow\) \(\triangle ASD\) is isosceles. The point \(K\) is the midpoint of \(AD\) , then \(SK\) is the height in the triangle \(\triangle ASD\) , and \(OK\) is the height in the triangle \(AOD\) \(\ Rightarrow\) plane \(SOK\) is perpendicular to the planes \(ASD\) and \(ABC\) \(\Rightarrow\) \(\angle SKO\) is a linear angle equal to the required dihedral angle.
In \(\triangle SKO\) : \(OK = \frac(1)(2)\cdot AB = \frac(1)(2)\cdot 10 = 5 = SO\)\(\Rightarrow\) \(\triangle SOK\) is isosceles right triangle\(\Rightarrow\) \(\angle SKO = 45^\circ\) .
Answer: 45
Task 7 #1855
Task level: More difficult than the exam
In the square \(ABCD\) : \(O\) is the intersection point of the diagonals; \(S\) is not in the plane of the square, \(SO \perp ABC\) . Find the angle between the planes \(ASD\) and \(BSC\) if \(SO = 5\) and \(AB = 10\) .
Right triangles \(\triangle SAO\) , \(\triangle SDO\) , \(\triangle SOB\) and \(\triangle SOC\) are equal in two sides and the angle between them (\(SO \perp ABC\) \(\Rightarrow\) \(\angle SOA = \angle SOD = \angle SOB = \angle SOC = 90^\circ\); \(AO = OD = OB = OC\) , because \(O\) is the point of intersection of the diagonals of the square, \(SO\) is the common side) \(\Rightarrow\) \(AS = DS = BS = CS\) \(\Rightarrow\) \(\triangle ASD\) and \(\triangle BSC\) are isosceles. The point \(K\) is the midpoint of \(AD\) , then \(SK\) is the height in the triangle \(\triangle ASD\) , and \(OK\) is the height in the triangle \(AOD\) \(\ Rightarrow\) the plane \(SOK\) is perpendicular to the plane \(ASD\) . The point \(L\) is the midpoint of \(BC\) , then \(SL\) is the height in the triangle \(\triangle BSC\) , and \(OL\) is the height in the triangle \(BOC\) \(\ Rightarrow\) the plane \(SOL\) (aka the plane \(SOK\) ) is perpendicular to the plane \(BSC\) . Thus, we obtain that \(\angle KSL\) is a linear angle equal to the desired dihedral angle.
\(KL = KO + OL = 2\cdot OL = AB = 10\)\(\Rightarrow\) \(OL = 5\) ; \(SK = SL\) - heights in equal isosceles triangles, which can be found using the Pythagorean theorem: \(SL^2 = SO^2 + OL^2 = 5^2 + 5^2 = 50\). It can be seen that \(SK^2 + SL^2 = 50 + 50 = 100 = KL^2\)\(\Rightarrow\) for a triangle \(\triangle KSL\) the inverse Pythagorean theorem holds \(\Rightarrow\) \(\triangle KSL\) is a right triangle \(\Rightarrow\) \(\angle KSL = 90^\ circ\) .
Answer: 90
Preparing students for the exam in mathematics, as a rule, begins with a repetition of the basic formulas, including those that allow you to determine the angle between the planes. Despite the fact that this section of geometry is covered in sufficient detail in the framework of school curriculum, many graduates need to repeat the basic material. Understanding how to find the angle between the planes, high school students will be able to quickly calculate the correct answer in the course of solving the problem and count on getting decent scores on the basis of the unified state exam.
Main nuances
So that the question of how to find the dihedral angle does not cause difficulties, we recommend that you follow the solution algorithm that will help you cope with the tasks of the exam.
First you need to determine the line along which the planes intersect.
Then on this line you need to choose a point and draw two perpendiculars to it.
The next step is finding trigonometric function dihedral angle, which is formed by perpendiculars. It is most convenient to do this with the help of the resulting triangle, of which the corner is a part.
The answer will be the value of the angle or its trigonometric function.
Preparation for the exam test together with Shkolkovo is the key to your success
During class the day before passing the exam many students are faced with the problem of finding definitions and formulas that allow you to calculate the angle between 2 planes. A school textbook is not always at hand exactly when it is needed. And in order to find the necessary formulas and examples of their correct application, including for finding the angle between planes on the Internet online, sometimes you need to spend a lot of time.
Mathematical portal "Shkolkovo" offers new approach to prepare for the state exam. Classes on our website will help students identify the most difficult sections for themselves and fill gaps in knowledge.
We have prepared and clearly stated all necessary material. Basic definitions and formulas are presented in the "Theoretical Reference" section.
In order to better assimilate the material, we also suggest practicing the corresponding exercises. A large selection of tasks of varying degrees of complexity, for example, on, is presented in the Catalog section. All tasks contain a detailed algorithm for finding the correct answer. The list of exercises on the site is constantly supplemented and updated.
Practicing in solving problems in which it is required to find the angle between two planes, students have the opportunity to save any task online to "Favorites". Thanks to this, they will be able to return to it the necessary number of times and discuss the course of its decision with school teacher or a tutor.
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Slides captions:
DOUBLE ANGLE Mathematics teacher GOU secondary school №10 Eremenko M.A.
The main objectives of the lesson: Introduce the concept of a dihedral angle and its linear angle Consider tasks for the application of these concepts
Definition: A dihedral angle is a figure formed by two half-planes with a common boundary line.
The value of a dihedral angle is the value of its linear angle. AF ⊥ CD BF ⊥ CD AFB is the linear angle of the dihedral angle ACD B
Let us prove that all linear angles of a dihedral angle are equal to each other. Consider two linear angles AOB and A 1 OB 1 . Rays OA and OA 1 lie on the same face and are perpendicular to OO 1, so they are co-directed. Rays OB and OB 1 are also co-directed. Therefore, ∠ AOB = ∠ A 1 OB 1 (as angles with codirectional sides).
Examples of dihedral angles:
Definition: The angle between two intersecting planes is the smallest of the dihedral angles formed by these planes.
Task 1: In the cube A ... D 1 find the angle between the planes ABC and CDD 1 . Answer: 90o.
Task 2: In the cube A ... D 1 find the angle between the planes ABC and CDA 1 . Answer: 45o.
Task 3: In the cube A ... D 1 find the angle between the planes ABC and BDD 1 . Answer: 90o.
Task 4: In the cube A ... D 1 find the angle between the planes ACC 1 and BDD 1 . Answer: 90o.
Task 5: In the cube A ... D 1 find the angle between the planes BC 1 D and BA 1 D . Solution: Let O be the midpoint of B D. A 1 OC 1 is the linear angle of the dihedral angle A 1 B D C 1 .
Problem 6: In the tetrahedron DABC all edges are equal, point M is the midpoint of edge AC. Prove that ∠ DMB is a linear angle of dihedral angle BACD .
Decision: Triangles ABC and ADC are regular, so BM ⊥ AC and DM ⊥ AC and hence ∠ DMB is the linear angle of the dihedral angle DACB .
Task 7: From the vertex B of the triangle ABC, the side AC of which lies in the plane α, a perpendicular BB 1 is drawn to this plane. Find the distance from point B to the line AC and to the plane αif AB=2, ∠BAC=150 0 and the dihedral angle BACB 1 is 45 0 .
Solution: ABC is an obtuse triangle with an obtuse angle A, so the base of height BK lies on the extension of side AC. VC is the distance from point B to AC. BB 1 - distance from point B to plane α
2) Since AS ⊥VK, then AS⊥KV 1 (according to the theorem, converse theorem about three perpendiculars). Therefore, ∠VKV 1 is the linear angle of the dihedral angle BACB 1 and ∠VKV 1 =45 0 . 3) ∆VAK: ∠A=30 0 , VK=VA sin 30 0 , VK =1. ∆VKV 1: VV 1 \u003d VK sin 45 0, VV 1 \u003d
Stereometry
Chapter 9
9.8. Dihedral angle and its linear angle
A plane is divided by a straight line lying in it into two half-planes.
Definition 1
The figure formed by two half-planes coming out of one straight line, together with the part of space bounded by these half-planes, is called a dihedral angle. Half-planes are called faces, and their common straight line is called an edge of a dihedral angle.
The faces of the dihedral angle divide the space into two regions: inner region given dihedral angle and its outer region.
Definition 2
Two dihedral angles are said to be equal if one of them can be combined with the other in such a way that their interior regions are aligned.
Definition 3
The angle between two perpendiculars to the edge of a dihedral angle, drawn in its faces from one point of the edge, is called the linear angle of the dihedral angle.
one . The angle () resulting from the intersection of a dihedral angle by a plane perpendicular to its edge is the linear angle of the given dihedral angle.
2. The value of the linear angle does not depend on the position of its vertex on the edge, i.e. .
3 . Linear angles of equal dihedral angles are equal (follows from Definitions 2 and 3).
Definition 4
Of the two dihedral angles, the one that has the larger (smaller) linear angle is called the larger (smaller) one. For units of measurement of dihedral angles, such dihedral angles are taken, the linear angles of which are equal to