\(\blacktriangleright\) Dihedral angle is an angle formed by two half-planes and a 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 (and spicy or straight) 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: carry out \(FG\perp \pi\) ;

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 right-angled.
Note also that the plane \(AFG\) constructed in this way is perpendicular to both planes \(\xi\) and \(\pi\) . Therefore, we can say it differently: 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 and \(\xi\ ) , and \(\pi\) .

Task 1 #2875

Task level: More difficult than the Unified State Exam

Given a 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 \(a\) . Consequently, all side faces are equal equilateral triangles. Let's find the angle between the faces \(SAD\) and \(SCD\) .

Let's do \(CH\perp SD\) . Because \(\triangle SAD=\triangle SCD\), then \(AH\) will also be the height of \(\triangle SAD\) . Therefore, by definition, \(\angle AHC=\alpha\) is the linear angle of the 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 \(a\), therefore, \(CH=AH=\frac(\sqrt3)2a\) .
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 Unified State 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 right angles, 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 a straight line \(a\), the line of intersection of \(\pi_2\) and \(\pi_3\) be a straight line \(b\), and the line of intersection \(\pi_3\) and \(\pi_1\) – 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”).

Let's mark the points \(A\in a, B\in b\) so that \(AB\perp a, AB\perp b\) (this is possible since \(a\parallel b\) ). Let us mark \(C\in c\) so that \(BC\perp c\) , therefore, \(BC\perp b\) . 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 therefore to any line from this plane, in particular, the line \ (AC\) .

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 Unified State Exam

Given straight 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 lines \(a\) and \(c\) and the plane formed by 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 straight lines cannot lie in the same plane. Let us mark the point \(A\) on the line \(a\) and draw \(AB\perp b\) and \(AC\perp c\) . Then \(\triangle AOB=\triangle AOC\) as rectangular along the hypotenuse and acute angle. Therefore, \(OB=OC\) and \(AB=AC\) .
Let's do \(AH\perp (BOC)\) . Then by the theorem about three perpendiculars \(HC\perp c\) , \(HB\perp b\) . Since \(AB=AC\) , then \(\triangle AHB=\triangle AHC\) as rectangular along the hypotenuse and leg. Therefore, \(HB=HC\) . This means that \(OH\) ​​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 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 angle. Since we chose the point \(A\) arbitrarily, 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 the 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 Unified State Exam

The planes \(\pi_1\) and \(\pi_2\) intersect along the straight line \(l\) on which the points \(M\) and \(N\) lie. The segments \(MA\) and \(MB\) are perpendicular to the straight 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, \(BN^2 = BM^2 + MN^2\), from which \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 Unified State 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 \((ABCD)\) , in addition, \(M\) is the point of intersection 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.

Let's 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 point of intersection of the diagonals of the square, then \(M\) is the middle of \(AC\), therefore, \(MN\) is the 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 theorem of three perpendiculars, \(A_1N\) 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 Unified State Exam

In a square \(ABCD\) : \(O\) – the point of intersection of the diagonals; \(S\) – does not lie 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 \(O\) – point of intersection of the diagonals of the square, \(SO\) – common side) \(\Rightarrow\) \(AS = SD\) \(\Rightarrow\) \(\triangle ASD\) – isosceles. Point \(K\) is the middle 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 planes \(ASD\) and \(ABC\) \(\Rightarrow\) \(\angle SKO\) – linear angle equal to the desired dihedral angle.

In \(\triangle SKO\) : \(OK = \frac(1)(2)\cdot AB = \frac(1)(2)\cdot 10 = 5 = SO\)\(\Rightarrow\) \(\triangle SOK\) – isosceles right triangle \(\Rightarrow\) \(\angle SKO = 45^\circ\) .

Answer: 45

Task 7 #1855

Task level: More difficult than the Unified State Exam

In a square \(ABCD\) : \(O\) – the point of intersection of the diagonals; \(S\) – does not lie 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 \(O\) – point of intersection of the diagonals of the square, \(SO\) – common side) \(\Rightarrow\) \(AS = DS = BS = CS\) \(\Rightarrow\) \(\triangle ASD\) and \(\triangle BSC\) are isosceles. Point \(K\) is the middle 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 plane \(ASD\) . Point \(L\) is the middle of \(BC\), then \(SL\) is the height in the triangle \(\triangle BSC\), and \(OL\) is the height in the triangle \(BOC\) \(\ Rightarrow\) plane \(SOL\) (aka 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 noticed that \(SK^2 + SL^2 = 50 + 50 = 100 = KL^2\)\(\Rightarrow\) for a triangle \(\triangle KSL\) the inverse Pythagorean theorem holds \(\Rightarrow\) \(\triangle KSL\) – right triangle \(\Rightarrow\) \(\angle KSL = 90^\ circ\) .

Answer: 90

Preparing students to take the Unified State Exam in mathematics, as a rule, begins with repeating basic formulas, including those that allow you to determine the angle between planes. Despite the fact that this section of geometry is covered in sufficient detail within the school curriculum, many graduates need to repeat the basic material. Understanding how to find the angle between planes, high school students will be able to quickly calculate the correct answer when solving a problem and count on receiving decent scores on the results of passing the unified state exam.

Main nuances

To ensure that the question of how to find a dihedral angle does not cause difficulties, we recommend following a solution algorithm that will help you cope with Unified State Examination tasks.

First you need to determine the straight line along which the planes intersect.

Then you need to select a point on this line and draw two perpendiculars to it.

The next step is to find the trigonometric function of the dihedral angle formed by the perpendiculars. The most convenient way to do this is with the help of the resulting triangle, of which the angle is a part.

The answer will be the value of the angle or its trigonometric function.

Preparing for the exam test with Shkolkovo is the key to your success

During classes on the eve of passing the Unified State Exam, many schoolchildren are faced with the problem of finding definitions and formulas that allow them to calculate the angle between 2 planes. A school textbook is not always at hand exactly when 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.

The Shkolkovo mathematical portal offers a new approach to preparing 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 presented all the necessary material. Basic definitions and formulas are presented in the “Theoretical Information” section.

In order to better understand the material, we also suggest practicing the appropriate exercises. A large selection of tasks of varying degrees of complexity, for example, on, is presented in the “Catalogue” section. All tasks contain a detailed algorithm for finding the correct answer. The list of exercises on the website is constantly supplemented and updated.

While practicing solving problems that require finding the angle between two planes, students have the opportunity to save any task online as “Favorites.” Thanks to this, they will be able to return to it the required number of times and discuss the progress of its solution with a school teacher or tutor.

To use presentation previews, create a Google account and log in to it: https://accounts.google.com

Slide captions:

DIHEDRAL ANGLE Mathematics teacher GOU secondary school No. 10 Eremenko M.A.

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 straight line.

The magnitude of a dihedral angle is the magnitude of its linear angle. AF ⊥ CD BF ⊥ CD AFB - linear dihedral angle ACD B

Let us prove that all linear angles of a dihedral angle are equal to each other. Let's 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 codirectional. Beams OB and OB 1 are also co-directed. Therefore, ∠ AOB = ∠ A 1 OB 1 (like angles with co-directed 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 cube A ... D 1, find the angle between planes ABC and CDD 1. Answer: 90 o.

Problem 2: In cube A ... D 1, find the angle between planes ABC and CDA 1. Answer: 45 o.

Problem 3: In cube A ... D 1, find the angle between planes ABC and BDD 1. Answer: 90 o.

Problem 4: In the cube A ... D 1, find the angle between the planes ACC 1 and BDD 1. Answer: 90 o.

Problem 5: In cube A ... D 1, find the angle between planes BC 1 D and BA 1 D. Solution: Let O be the midpoint of B D. A 1 OC 1 – 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 middle of edge AC. Prove that ∠ DMB is the linear angle of the dihedral angle BACD .

Solution: Triangles ABC and ADC are regular, therefore, BM ⊥ AC and DM ⊥ AC and hence ∠ DMB is the linear angle of dihedral angle DACB.

Problem 7: From vertex B of triangle ABC, 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 straight line AC and to the plane α, if AB=2, ∠ВАС=150 0 and the dihedral angle ВАСВ 1 is equal to 45 0.

Solution: ABC is an obtuse triangle with obtuse angle A, therefore the base of the altitude BC lies on the extension of side AC. VC – distance from point B to AC. BB 1 – distance from point B to plane α

2) Since AC ⊥BK, then AC⊥KB 1 (by the theorem inverse to the theorem about three perpendiculars). Therefore, ∠VKV 1 is the linear angle of the dihedral angle BASV 1 and ∠VKV 1 =45 0 . 3) ∆VAK: ∠A=30 0, VK=VA·sin 30 0, VK =1. ∆ВКВ 1: ВВ 1 =ВК· sin 45 0 , ВВ 1 =

Stereometry

Chapter 9. Lines and planes in space

9.8. Dihedral angle and its linear angle

A plane is divided by a line lying in it into two half-planes.

Definition 1

A figure formed by two half-planes emerging from one straight line, together with the part of space limited 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 a dihedral angle divide space into two regions: the internal region of a given dihedral angle and its external region.

Definition 2

Two dihedral angles are said to be equal if one of them can be combined with the other so that their internal regions are aligned.

Definition 3

The angle between two perpendiculars to the edge of a dihedral angle, drawn at its faces from one point on the edge, is called the linear angle of the dihedral angle.

1 . The angle () obtained when a dihedral angle is intersected by a plane perpendicular to its edge is the linear angle of the given dihedral angle.

2. The magnitude 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 two dihedral angles, the one that has the larger (smaller) linear angle is called the larger (smaller) one. The units of measurement for dihedral angles are those dihedral angles whose linear angles are equal

In geometry, two important characteristics are used to study figures: the lengths of the sides and the angles between them. In the case of spatial figures, dihedral angles are added to these characteristics. Let's look at what it is, and also describe the method for determining these angles using the example of a pyramid.

The concept of dihedral angle

Everyone knows that two intersecting lines form a certain angle with the vertex at the point of their intersection. This angle can be measured using a protractor or you can use trigonometric functions to calculate it. An angle formed by two right angles is called linear.

Now imagine that in three-dimensional space there are two planes that intersect in a straight line. They are shown in the picture.

A dihedral angle is the angle between two intersecting planes. Just like linear, it is measured in degrees or radians. If to any point on the line along which the planes intersect, we restore two perpendiculars lying in these planes, then the angle between them will be the desired dihedral. The easiest way to determine this angle is to use the equations of planes in general form.

Equation of planes and formula for the angle between them

The equation of any plane in space is generally written as follows:

A × x + B × y + C × z + D = 0.

Here x, y, z are the coordinates of points belonging to the plane, the coefficients A, B, C, D are some known numbers. The convenience of this equality for calculating dihedral angles is that it explicitly contains the coordinates of the direction vector of the plane. We will denote it n¯. Then:

Vector n¯ is perpendicular to the plane. The angle between two planes is equal to the angle between their n 1 ¯ and n 2 ¯. It is known from mathematics that the angle formed by two vectors is uniquely determined from their scalar product. This allows us to write a formula for calculating the dihedral angle between two planes:

φ = arccos (|(n 1 ¯ × n 2 ¯)| / (|n 1 ¯| × |n 2 ¯|)).

If we substitute the coordinates of the vectors, the formula will be written explicitly:

φ = arccos (|A 1 × A 2 + B 1 × B 2 + C 1 × C 2 | / (√(A 1 2 + B 1 2 + C 1 2) × √(A 2 2 + B 2 2 + C 2 2))).

The modulus sign in the numerator is used to define only the acute angle, since the dihedral angle is always less than or equal to 90 o.

Pyramid and its corners

A pyramid is a figure that is formed by one n-gon and n triangles. Here n is an integer equal to the number of sides of the polygon that is the base of the pyramid. This spatial figure is a polyhedron or polyhedron, since it consists of flat faces (sides).

Pyramid polyhedrons can be of two types:

• between the base and the side (triangle);
• between the two sides.

If we are considering a regular pyramid, then the named angles for it are not difficult to determine. To do this, using the coordinates of three known points, you should create an equation of the planes, and then use the formula given in the paragraph above for the angle φ.

Below we give an example in which we show how to find dihedral angles at the base of a regular quadrangular pyramid.

Quadrangular and the angle at its base

Let us assume that we are given a regular pyramid with a square base. The length of the side of the square is a, the height of the figure is h. Let's find the angle between the base of the pyramid and its side.

Let's place the origin of the coordinate system at the center of the square. Then the coordinates of points A, B, C, D shown in the figure will be equal:

A = (a/2; -a/2; 0);

B = (a/2; a/2; 0);

C = (-a/2; a/2; 0);

Let's consider the planes ACB and ADB. Obviously, the direction vector n 1 ¯ for the plane ACB will be equal to:

To determine the direction vector n 2 ¯ of the ADB plane, we proceed as follows: we find arbitrary two vectors that belong to it, for example, AD¯ and AB¯, then we calculate their vector product. Its result will give the coordinates n 2 ¯. We have:

AD¯ = D - A = (0; 0; h) - (a/2; -a/2; 0) = (-a/2; a/2; h);

AB¯ = B - A = (a/2; a/2; 0) - (a/2; -a/2; 0) = (0; a; 0);

n 2 ¯ = = [(-a/2; a/2; h) × (0; a; 0)] = (-a × h; 0; -a 2 /2).

Since multiplying and dividing a vector by a number does not change its direction, we transform the resulting n 2 ¯ by dividing its coordinates by -a, we get:

We have defined the direction vectors n 1 ¯ and n 2 ¯ for the base planes ACB and side plane ADB. It remains to use the formula for the angle φ:

φ = arccos (|(n 1 ¯ × n 2 ¯)| / (|n 1 ¯| × |n 2 ¯|)) = arccos (a / (2 × √h 2 + a 2 /4)).

Let's transform the resulting expression and rewrite it like this:

φ = arccos (a / √(a 2 + 4 × h 2)).

We have obtained a formula for the dihedral angle at the base for a regular quadrangular pyramid. Knowing the height of the figure and the length of its side, you can calculate the angle φ. For example, for the Cheops pyramid, the base side of which is 230.4 meters, and the initial height was 146.5 meters, the angle φ will be equal to 51.8 o.

You can also determine the dihedral angle for a quadrangular regular pyramid using the geometric method. To do this, it is enough to consider a right triangle formed by the height h, half the length of the base a/2 and the apothem of an isosceles triangle.

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