Watch what is "angle" in other dictionaries. Corners with heated parties
This material is devoted to such a concept as the angle between two intersecting straight. In the first point we will explain what he is, and show it on the illustrations. Then we will analyze how the sinus can be found, the cosine of this angle and the angle itself (separately consider cases with the plane and three-dimensional space), we give the necessary formulas and show the examples on the examples, how exactly they are applied in practice.
In order to understand what an angle is formed by the intersection of two direct, we will need to recall the determination of the angle, perpendicularity and the intersection points.
Definition 1.
We call two straight intersecting if they have one common point. This point is called the intersection point of two straight lines.
Each direct is separated by the point of intersection on the rays. Both direct at the same time form 4 corners, of which two are vertical, and two are adjacent. If we know the measure of one of them, we can identify other remaining.
Suppose we know that one of the corners is equal to α. In this case, an angle that is vertical in relation to it will also be equal to α. To find the remaining angles, we need to calculate the difference between 180 ° - α. If α is equal to 90 degrees, then all the angles will be straight. Intersecting at the right corner of the line are called perpendicular (individual article is devoted to the concept of perpendicularity).
Take a look at the drawing:
Let us turn to the formulation of the basic definition.
Definition 2.
The angle formed by two intersecting straight is a measure of a smaller of the 4-corners, which form two of these straight.
From the definition, it is necessary to make an important conclusion: the angle size in this case will be expressed by any real number in the interval (0, 90]. If direct is perpendicular, then the angle between them will be equal to 90 degrees.
The ability to find a measure of the angle between two intersecting direct is useful for solving many practical tasks. The solution method can be selected from several options.
For a start, we can take geometric methods. If we know something about additional corners, then you can tie them with the angle we need using the properties of equal or similar shapes. For example, if we know the side of the triangle and you need to calculate the angle between the direct, on which these parties are located, then for solutions, the cosine theorem is suitable. If we have a rectangular triangle, then for calculations, we also use the knowledge of sinus, cosine and tangent angle.
The coordinate method is also very convenient for solving problems of this type. Let us explain how to use it correctly.
We have a rectangular (decartian) coordinate system O x Y, in which two straight lines are given. Denote them with letters a and b. Direct with this can be described using any equations. The source straight lines have the intersection point M. How to determine the desired angle (denote it α) between these straight?
Let's start with the wording of the basic principle of finding an angle under specified conditions.
We know that with the concept of straight line, such concepts as a guide and normal vector are closely connected. If we have an equation to some straight, you can take the coordinates of these vectors from it. We can do it immediately for two intersecting straight lines.
The angle formed by two intersecting straight, can be found using:
- the angle between the guide vectors;
- angle between normal vectors;
- the angle between the normal vector is one straight and e-guide vector.
Now consider each way separately.
1. Suppose that we have straight A with the guide vector A → \u003d (A X, A Y) and straight b with the guide vector B → (B x, b y). Now postpone two vectors A → and B → from the intersection point. After that, we will see that they will be located each on their straight. Then we have four options for their mutual location. See illustration:
If the angle between two vectors is not stupid, then it will be the angle we need to go between intersecting straight a and b. If it is stupid, then the desired angle will be equal to the corner adjacent to the angle of A →, B → ^. Thus, α \u003d a →, b → ^ if A →, B → ^ ≤ 90 °, and α \u003d 180 ° - A →, B → ^, if A →, B → ^\u003e 90 °.
Based on the fact that the cosines of equal angles are equal, we can rewrite the resulting equality: cos α \u003d cos a →, b → ^, if A →, B → ^ ≤ 90 °; COS α \u003d COS 180 ° - A →, B → ^ \u003d - COS A →, B → ^, if A →, B → ^\u003e 90 °.
In the second case, the formulas were used. In this way,
cos α cos a →, b → ^, cos a →, b → ^ ≥ 0 - cos a →, b → ^, cos a →, b → ^< 0 ⇔ cos α = cos a → , b → ^
We write the last formula with the words:
Definition 3.
The cosine of an angle formed by two intersecting straight, will be equal to the module of the cosine of the angle between its guide vectors.
The general appearance of the cosine formula of the angle between two vectors A → \u003d (a x, a y) and B → \u003d (b x, b y) looks like this:
cOS A →, B → ^ \u003d A →, B → ^ A → · B → \u003d a x · b x + a y + b y a x 2 + a y 2 · b x 2 + b y 2
From it we can derive the cosine formula of the angle between the two specified direct:
cos α \u003d a x · b x + a y + b y a x 2 + a y 2 · b x 2 + b y 2 \u003d a x · b x + a y + b y a x 2 + a y 2 · b x 2 + b y 2
Then the angle itself can be found on the following formula:
α \u003d a r c cos a x · b x + a y + b y a x 2 + a y 2 · b x 2 + b y 2
Here a → \u003d (a x, a y) and B → \u003d (b x, b y) are the guide vectors of the specified direct.
Let us give an example of solving the problem.
Example 1.
In the rectangular coordinate system on the plane, two intersecting straight lines A and b are given. They can be described by parametric equations x \u003d 1 + 4 · λ y \u003d 2 + λ λ ∈ R and x 5 \u003d y - 6 - 3. Calculate the angle between these straight.
Decision
In our condition, there is a parametric equation, it means that for this straight, we can immediately write the coordinates of its guide vector. For this, we need to take the values \u200b\u200bof the coefficients when parameter, i.e. Direct x \u003d 1 + 4 · λ y \u003d 2 + λ λ ∈ R will have a guide vector A → \u003d (4, 1).
The second direct is described using the canonical equation x 5 \u003d y - 6 - 3. Here we can take the coordinates from the denominators. Thus, this direct has a guide vector B → \u003d (5, - 3).
Next, go directly to finding the angle. To do this, we simply substitute the available coordinates of the two vectors in the above formorm α \u003d a r c cos a x · b x + a y + b y a x 2 + a y 2 · b x 2 + b y 2. We get the following:
α \u003d a r c cos 4 · 5 + 1 · (- 3) 4 2 + 1 2 · 5 2 + (- 3) 2 \u003d a r c cos 17 17 · 34 \u003d a r c cos 1 2 \u003d 45 °
Answer: Data direct form an angle of 45 degrees.
We can solve such a task by finding the angle between normal vectors. If we have straight A with a normal Na → \u003d (NAX, NAX) vector and straight b with a normal NB → \u003d (NBX, NBY) vector, then the angle between them will be equal to the corner between Na → and Nb → either the corner that will be adjacent to Na →, Nb → ^. This method is shown in the picture:
Formulas for calculating the cosine of the angle between the intersecting straight and most of this angle with the help of coordinates of normal vectors look like this:
cos α \u003d cos n a →, n b → ^ \u003d n И x · n b x + n a y + n b y n a x 2 + n b y 2 · n b x 2 + n b y 2 α \u003d a r c cos n a x · n b x + n a y + n b y n a x 2 + n a y 2 · n b x 2 + n b y 2
Here n A → and N B → denote the normal vectors of the two set direct.
Example 2.
In the rectangular coordinate system, two straight lines are given using equations 3 x + 5 y - 30 \u003d 0 and x + 4 y - 17 \u003d 0. Find the sinus, cosine angle between them and the magnitude of this corner itself.
Decision
The source straight lines are given using normal equations of the direct form A x + b y + c \u003d 0. Normal vector Denote by n → \u003d (a, b). We will find the coordinates of the first normal vector for one straight and write them: n a → \u003d (3, 5). For the second direct x + 4 y - 17 \u003d 0, the normal vector will have coordinates N B → \u003d (1, 4). Now add the obtained values \u200b\u200bin the formula and calculate the result:
cos α \u003d cos n a →, n b → ^ \u003d 3 · 1 + 5 · 4 3 2 + 5 2 · 1 2 + 4 2 \u003d 23 34 · 17 \u003d 23 2 34
If we are known to cosine angle, then we can calculate it sinus using a basic trigonometric identity. Since the angle α, formed by straight, is not blunt, then sin α \u003d 1 - cos 2 α \u003d 1 - 23 2 34 2 \u003d 7 2 34.
In this case, α \u003d a r c cos 23 2 34 \u003d a r c sin 7 2 34.
Answer: cos α \u003d 23 2 34, sin α \u003d 7 2 34, α \u003d a r c cos 23 2 34 \u003d a r c sin 7 2 34
We will analyze the last case - finding the angle between straight, if we know the coordinates of the guide vector of one straight and normal vector of another.
Suppose that direct A has a guide vector A → \u003d (A x, A Y), and the straight line B is the normal vector n B → \u003d (n b x, n b y). We need to postpone these vectors from the intersection point and consider all the options for their mutual location. See in Picture:
If the value of the angle between the specified vectors is not more than 90 degrees, it turns out that it will complement the angle between A and B to the direct angle.
a →, N B → ^ \u003d 90 ° - α if A →, N B → ^ ≤ 90 °.
If it is less than 90 degrees, then we will get the following:
a →, N B → ^\u003e 90 °, then A →, N B → ^ \u003d 90 ° + α
Using the rule of equal cosine equal angles, write:
cos a →, n b → ^ \u003d cos (90 ° - α) \u003d sin α at a →, n b → ^ ≤ 90 °.
cOS A →, N B → ^ \u003d COS 90 ° + α \u003d - sin α at a →, n b → ^\u003e 90 °.
In this way,
sin α \u003d cos a →, nb → ^, a →, nb → ^ ≤ 90 ° - cos a →, nb → ^, a →, nb → ^\u003e 90 ° ⇔ sin α \u003d cos a →, nb → ^, A →, NB → ^\u003e 0 - COS A →, NB → ^, A →, NB → ^< 0 ⇔ ⇔ sin α = cos a → , n b → ^
We formulate output.
Definition 4.
To find the sine angle between two straight lines intersecting on the plane, you need to calculate the cosine module between the guide vector of the first straight and normal vector of the second.
We write the necessary formulas. Finding the sine corner:
sin α \u003d cos a →, n b → ^ \u003d a x · n b x + a y · n b y a x 2 + a y 2 · n b x 2 + n b y 2
Finding the corner:
α \u003d a r c sin \u003d a x · n b x + a y · n b y a x 2 + a y 2 · n b x 2 + n b y 2
Here a → is the first line guide vector, and N B → is a normal second vector.
Example 3.
Two intersecting straight lines are set by the equations x - 5 \u003d y - 6 3 and x + 4 y - 17 \u003d 0. Find the crossing angle.
Decision
We take the coordinates of the guide and normal vector from the specified equations. It turns out a → \u003d (- 5, 3) and n → b \u003d (1, 4). We take the formula α \u003d a r c sin \u003d a x · n b x + a y · n b y a x 2 + a y 2 · n b x 2 + n b y 2 and consider:
α \u003d a r c sin \u003d - 5 · 1 + 3 · 4 (- 5) 2 + 3 2 · 1 2 + 4 2 \u003d a r c sin 7 2 34
Please note that we have taken equations from the previous task and got exactly the same result, but in another way.
Answer: α \u003d a r c sin 7 2 34
We give another way to find the desired angle using the angular coefficients of the specified direct.
We have direct A, which is given in the rectangular coordinate system using the Y \u003d k 1 · X + B 1 equation, and straight b, given as Y \u003d k 2 · X + B 2. These are equations direct with an angular coefficient. To find an angle of intersection, we use the formula:
α \u003d a r c cos k 1 · k 2 + 1 k 1 2 + 1 · k 2 2 + 1, where k 1 and k 2 are the angular coefficients of the specified direct. To obtain this entry, the formulas for determining the angle through the coordinates of normal vectors were used.
Example 4.
There are two straight-intersecting on the plane, given by the equations y \u003d - 3 5 x + 6 and y \u003d - 1 4 x + 17 4. Calculate the magnitude of the intersection angle.
Decision
The angular coefficients of our lines are equal to k 1 \u003d - 3 5 and k 2 \u003d - 1 4. We add them to the formula α \u003d a r c cos k 1 · k 2 + 1 k 1 2 + 1 · k 2 2 + 1 and we calculate:
α \u003d a r c cos - 3 5 · - 1 4 + 1 - 3 5 2 + 1 · - 1 4 2 + 1 \u003d a r c cos 23 20 34 24 · 17 16 \u003d a r c cos 23 2 34
Answer: α \u003d a r c cos 23 2 34
In the conclusions of this item, it should be noted that the formulas given here are not necessarily learning by heart. To do this, it is enough to know the coordinates of the guide and / or normal vectors of the specified direct and be able to determine them in different types of equations. But the formula for calculating the cosine of the angle is better remembered or recorded.
How to calculate the angle between intersecting straight in space
The calculation of such angle can be reduced to calculating the coordinates of the guide vectors and the determination of the angle formed by these vectors. For such examples, the same arguments that we have led to it are used.
Suppose that we have a rectangular coordinate system located in three-dimensional space. It contains two straight lines a and b with a point of intersection m. To calculate the coordinates of the guide vectors, we need to know the equations of these direct. Denote the guide vectors A → \u003d (a x, a y, a z) and b → \u003d (b x, b y, b z). To calculate the cosine of the angle between them, we use the formula:
cos α \u003d cos a →, b → ^ \u003d a →, b → a → · b → \u003d a x · b x + a y · b y + a z · b z a x 2 + a y 2 + a z 2 · b x 2 + b y 2 + b z 2
To find the corner itself, we will need this formula:
α \u003d a r c cos a x · b x + a y · b y + a z · b z a x 2 + a y 2 + a z 2 · b x 2 + b y 2 + b z 2
Example 5.
We have a straight line, given in three-dimensional space using an equation x 1 \u003d y - 3 \u003d z + 3 - 2. It is known that it intersects with the O Z axis. Calculate the angle of intersection and cosine of this angle.
Decision
Denote the angle that must be calculated, the letter α. We write the coordinates of the guide vector for the first direct - A → \u003d (1, - 3, - 2). For the appliqué axis, we can take the coordinate vector K → \u003d (0, 0, 1) as a guide. We received the necessary data and can add them to the desired formula:
cos α \u003d cos a →, k → ^ \u003d a →, k → a → · k → \u003d 1 · 0 - 3 · 0 - 2 · 1 1 2 + (- 3) 2 + (- 2) 2 · 0 2 + 0 2 + 1 2 \u003d 2 8 \u003d 1 2
As a result, we obtained that the angle we need will be equal to a r c cos 1 2 \u003d 45 °.
Answer: COS α \u003d 1 2, α \u003d 45 °.
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In this lesson, we will give the definition of the heated rays and prove the theorem about the equality of angles with the heated parties. Next, we will give the definition of the angle between intersecting straight and crossing straight. Consider what could be the angle between two straight. At the end of the lesson, we decide several tasks to find the corners between cross-lived straight.
Subject: Parallelism of straight and planes
Lesson: angles with air cooled sides. The angle between two straight
Any direct, for example Oo 1. (Fig. 1.), dissect plane into two half-planes. If the rays OA and O 1 A 1 parallel and lie in one half-plane, then they are called sonated.
Rays O 2 A 2 and OA are not co-controlled (Fig. 1.). They are parallel, but do not lie in one half-plane.
If the sides of two angles are cooled, then such angles are equal.
Evidence
Let us give parallel rays OA and O 1 A 1 and parallel rays OV and O 1 in 1 (Fig. 2.). That is, we have two angle Aah and A 1 O 1 in 1whose parties lie on the heated rays. We prove that these corners are equal.
On the side of the beam OA and O 1 A 1 Choose Points BUT and A 1. so that the segments OA and O 1 A 1 were equal. Similarly, a point IN and IN 1 choose so that the segments OV and O 1 in 1were equal.
Consider a quadrangle A 1 O 1 OA (Fig. 3.) OA and O 1 A 1 A 1 O 1 OA A 1 O 1 OA Oo 1. and AA 1. Parallel and equal.
Consider a quadrangle In 1 o 1 s. In this quadrangles OV and O 1 in 1 Parallel and equal. On the basis of the parallelogram, quadrangle In 1 o 1 s It is a parallelogram. As In 1 o 1 s - parallelogram, then Oo 1. and BB 1. Parallel and equal.
And straight AA 1. Parallel direct Oo 1., and straight BB 1. Parallel direct Oo 1.So direct AA 1. and BB 1. Parallel.
Consider a quadrangle 1 A 1 AV. In this quadrangles AA 1. and BB 1. Parallel and equal. On the basis of the parallelogram, quadrangle 1 A 1 AV It is a parallelogram. As 1 A 1 AV - parallelogram, then AU and 1 in 1 Parallel and equal.
Consider triangles Aah and A 1 O 1 in 1.Parties OA and O 1 A 1equal to construction. Parties OV and O 1 in 1also equal to construction. And how we proved and the parties AU and 1 in 1 Also equal. So triangles Aah and A 1 O 1 in 1equal in three sides. In equal triangles, equal angles lie against equal parties. So angles Aah and A 1 O 1 in 1equal to what was required to prove.
1) intersecting straight.
If direct intersecting, then we have four different angle. Angle between two straight, called the smallest corner between two straight. The angle between intersecting straight but and b. Denote by α (Fig. 4.). The angle α is such that.
Fig. 4. The angle between two intersecting straight
2) cross-lived straight
Let live but and b. Crossing. Choose an arbitrary point ABOUT. Through the point ABOUT Let's spend straight a 1., parallel to direct but, and straight b 1., parallel to direct b. (Fig. 5.). Straight a 1. and b 1. intersect at point ABOUT. The angle between two intersecting straight a 1. and b 1. , Corner φ, and is called an angle between cross-lived straight.
Fig. 5. The angle between two cross-country straight
Does the corner of the selected point depend on? Choose a point O 1.. Through the point O 1. Let's spend straight a 2., parallel to direct but, and straight b 2., parallel to direct b. (Fig. 6.). The angle between intersecting straight a 2. and b 2. Denote Φ 1.. Then angles φ and φ 1 -corners with heated parties. As we proved, such angles are equal to each other. It means that the magnitude of the angle between cross-country direct does not depend on the choice of point ABOUT.
Straight OV and Cd. parallel OA and Cd. Crossed. Find the angle between straight OA and Cd., if a:
1) ∠Aah \u003d 40 °.
Choose a point FROM. Pass through it Cd.. Let's spend Ca 1. parallel OA (Fig. 7.). Then corner A 1 cd. - the angle between crossing straight OA and Cd.. By the corners theorem with the heated parties, the angle A 1 cd. equal to the corner Aah, that is, 40 °.
Fig. 7. Find the angle between two straight
2) ∠Aah \u003d 135 °.
Let's do the same construction (Fig. 8.). Then the angle between cross-country straight OA and Cd. equal to 45 °, as it is the smallest of the corners that are obtained when crossing the direct Cd. and Ca 1..
3) ∠Aah \u003d 90 °.
Let's do the same construction (Fig. 9.). Then all the angles that are obtained when crossing direct Cd. and Ca 1. 90 ° are equal. The desired angle is 90 °.
1) Prove that the mid-sides of the spatial quadrilateral are vertices of the parallelogram.
Evidence
Let us give a spatial quadrangle Abcd.. M,N,K,L. - Middle Ribs BD,Ad,AC,BC. Accordingly (Fig. 10.). Need to prove that MNKL - parallelogram.
Consider a triangle AVD. Mn. Mn. Parallel AU And equals her half.
Consider a triangle ABC. LD - middle line. By the properties of the midline, LD Parallel AU And equals her half.
AND Mn., I. LD Parallel AU. It means Mn. Parallel LD By the theorem on three parallel straight lines.
We get that in the quadrangle MNKL - Party Mn. and LD parallel and equal because Mn. and LD equal half AU. So, on the basis of the parallelogram, quadrangle MNKL - parallelogram, which was required to prove.
2) Find the angle between straight AU and Cd.if corner Mnk \u003d 135 °.
As we have proven Mn. Parallel direct AU. NK - the middle line of the triangle ACD, according to the property, NK Parallel DC. So, through the point N. Pass two straight lines Mn. and NKwhich are parallel to cross-direct direct AU and DC respectively. So, the angle between straight Mn. and NK is an angle between cross-country straight AU and DC. We are given a stupid angle Mnk \u003d 135 °. The angle between straight Mn. and NK - The smallest of the corners obtained with the intersection of these direct, that is, 45 °.
So, we reviewed the angles with the heated parties and proved their equality. The angles between intersecting and crossing straight and solved several tasks to find the angle between two directs. In the next lesson, we will continue to solve problems and repeat the theory.
1. Geometry. 10-11 Class: Textbook for students of general educational institutions (basic and profile levels) / I. M. Smirnova, V. A. Smirnov. - 5th edition, revised and supplemented - M.: Mnemozina, 2008. - 288 p. : IL.
2. Geometry. 10-11 Class: Textbook for general educational institutions / Sharygin I. F. - M.: Drop, 1999. - 208 p.: Il.
3. Geometry. Grade 10: Textbook for general education institutions with in-depth and profile study of mathematics / e. V. Potoskuev, L. I. Zvalich. - 6th edition, stereotype. - M.: Drop, 008. - 233 p. : IL.
IN) BC. and D. 1 IN 1.
Fig. 11. Find the angle between straight
4. Geometry. 10-11 Class: Textbook for students of general educational institutions (basic and profile levels) / I. M. Smirnova, V. A. Smirnov. - 5th edition, corrected and supplemented - M.: Mnemozina, 2008. - 288 p.: Il.
Tasks 13, 14, 15 p. 54
Consisting of two different rays coming out of one point. Rays called. sides of W., and their overall start - vertex W. Let [ V.),[ Sun.) -
The side of the corner IN - His vertex - a plane determined by the sides of W. Figure divides the plane into two figures Figure 
i \u003d\u003d L, 2, also called. U. or flat corner, called. Internal area of \u200b\u200bflat U.
Two angles called. Equal (or congruent), if they can be combined so that their respective parties and vertices coincide. From any beam on the plane in this direction, the only U. can be postponed from it, equal to this W. Comparison of W. is carried out in two ways. If W. is considered as a pair of rays with a general start, to clarify some of the two W. More, it is necessary to combine in one plane of the vertex of the W. and one pair of them (see Fig. 1). If the second side of one W. will be located inside another W., then they say that the first W. is less than the second. The second way of comparison of U. is based on a comparison of each W. Some Nome. Equal W. will correspond to the same degrees or (see below), larger y. - more, smaller than.
Two U. Naz. adjacent if they have a total vertex and one side, and the other two sides form a straight line (see Fig. 2). Generally, W., having a total vertex and one common side, called. Digitious. W. Naz. Vertical, if the sides of one are continuing over the top of the sides of the other W. Vertical W. Are equal to each other. W., in the onward side form a straight, called. Expanded. Half expanded U. Naz. Direct W. Direct U. can be equivalent to otherwise: W., equal to its adjacent, called. straight. Inner flat U., not exceeding the deployed, is a convex area on the plane. For a unit of measurement of W. The 90th share of direct U., Naz. Degree.
Used, etc. Measure U. The numerical value of the radian measure of the U. is equal to the length of the arc, carved by the parties from the unit circumference. One radian is attributed to W., the corresponding arc, which is equal to its radius. Deployed W. is equal to radians.
When crossing two straight lies lying in the same plane, the third straight line is formed by U. (see Fig. 3): 1 and 5, 2 and 6, 4 and 8, s and 7th - Naz. respectively; 2 and 5, 3 and 8 - internal one-sided; 1 and 6, 4 and 7 - external one-sided; 3 and 5, 2 and 8 - internal climb lying; 1 and 7, 4 and 6 - external passages lying.
In practice. Tasks are advisable to consider W. How to measure the rotation of the fixed beam around it starts to a given position. Depending on the direction of the turn of W. In this case, both positive and negative are considered. Thus, U. in this sense can have any value of any. W. How the turn of the ray is considered in the theory of trigonometric. Functions: For any values \u200b\u200bof the argument (U.), you can define trigonometric values. functions. The concept of W. in geometrich. The system, the basis of the K-Roy is the point-and-vector axiomatics, the radar differs from the definitions of W. As the figures - in this axiomatics under W. understand a certain metric. The magnitude associated with two vectors using the scalar multiplication operations operation. It is, each pair of vectors of Ai Baets a certain angle - the number associated with the vector formula
where ( a, B.) -
Scalar product of vectors.
The concept of W. as a flat figure and as some numerical size is used in various geometrich. Tasks, in-ry u. Defined in a special way. So, under W. Between intersecting curves having certain tangents at the intersection point, U., formed by these tangents.
The corner between the straight and plane is accepted by U., formed by the straight and its rectangular projection on the plane; It is measured ranging from 0
Mathematical encyclopedia. - M.: Soviet Encyclopedia. I. M. Vinogradov. 1977-1985.
Synonyms:Watch what is "angle" in other dictionaries:
corner - Angle / Ek / ... Morphemno-spell dictionary
Husband. Fracture, breakdown, knee, elbow, protrusion or hall (vydina) about one face. The corner is linear, all sorts of two oncoming traits and the interval of them; an angle of plane or in planes, meeting two planes or walls; Corner thick, body, meeting in one ... Explanatory dictionary of Daly
Corner, Angle, on (c) corner and (mat.) In the corner, m. 1. Part of the plane between two straight lines emanating from one point (mat.). The top of the corner. The side of the corner. Measurement of the angle of degrees. Right angle. (90 °). Sharp corner. (less than 90 °). Obtuse angle.… … Explanatory Dictionary Ushakov
ANGLE - (1) Attack the angle between the direction of the air flow, which bears on the wing of the aircraft, and the chord of the sections of the wing. From this angle depends the value of the lifting force. The angle in which the lifting force is maximum, is called a critical angle of attack. At ... ... Large polytechnic encyclopedia
- (flat) geometric shape formed by two rays (corner sides) extending from one point (the peak of the angle). Any angle with a vertex in the center of some circumference (central angle) determines on the circumference of an arc of AV, limited points ... ... Big Encyclopedic Dictionary
The head of the corner, because of the angle, the bear corner, the bad corner, in all corners .. the dictionary of Russian synonyms and similar expressions in the meaning of expressions. under. ed. N. Abramova, M.: Russian Dictionaries, 1999. The angle of the vertex, the angular point; Dereleng, Rubbear, Ninetina, Rumbers, ... ... Synonym dictionary
angle - angle, genus. corner; Run. about the corner, in (for) the corner and in the speech of mathematicians in the corner; MN. Corners, genus. Corners. In the proposed and sustainable combinations: for the angle and permissible for the angle (go, wrap, etc.), from the angle to the angle (move, be located, etc.), the angle ... ... Dictionary of the difficulties of pronunciation and stress in modern Russian
Angle, corner, about the corner, on (c) corner, husband. 1. (in the corner.). In geometry: a flat figure formed by two rays (in 3 meanings), outgoing from one point. The top of the corner. Straight y. (90 °). Acute. (less than 90 °). Stupid y. (more than 90 °). External and internal ... ... Explanatory dictionary of Ozhegov
angle - angle, angle, m. A quarter bet, when declaring which the edge of the map is bent. ◘ Ace and lady peak with an angle // killed. A.I. Polyzhaev. Day in Moscow, 1832. He is scattered after lunch. Ponteps crack decks, ... ... Card terminology and zagon XIX century
