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The construction of an angle equal to a given one is brief. Constructing an angle equal to a given one

Lesson objectives:

  • Formation of the ability to analyze the studied material and the skills of applying it to solve problems;
  • Show the significance of the concepts being studied;
  • Development cognitive activity and independence in acquiring knowledge;
  • Cultivating interest in the subject and a sense of beauty.


Lesson objectives:

  • Develop skills in constructing an angle equal to a given one using a scale ruler, compass, protractor and drawing triangle.
  • Test students' problem-solving skills.

Lesson plan:

  1. Repetition.
  2. Constructing an angle equal to a given one.
  3. Analysis.
  4. Construction example first.
  5. Construction example two.

Repetition.

Corner.

Flat angle- an unlimited geometric figure formed by two rays (sides of an angle) emerging from one point (vertex of the angle).

An angle is also called a figure formed by all points of the plane enclosed between these rays (Generally speaking, two such rays correspond to two angles, since they divide the plane into two parts. One of these angles is conventionally called internal, and the other - external.
Sometimes, for brevity, the angle is called the angular measure.

There is a generally accepted symbol to denote an angle: , proposed in 1634 by the French mathematician Pierre Erigon.

Corner is a geometric figure (Fig. 1), formed by two rays OA and OB (sides of the angle), emanating from one point O (vertex of the angle).

An angle is denoted by a symbol and three letters indicating the ends of the rays and the vertex of the angle: AOB (and the letter of the vertex is the middle one). Angles are measured by the amount of rotation of ray OA around vertex O until ray OA moves to position OB. There are two widely used units for measuring angles: radians and degrees. For radian measurement of angles, see below in the paragraph “Arc Length”, as well as in the chapter “Trigonometry”.

Degree system for measuring angles.

Here the unit of measurement is a degree (its designation is °) - this is a rotation of the beam by 1/360 of a full revolution. Thus, full turn beam is equal to 360 o. One degree is divided into 60 minutes (symbol ‘); one minute – respectively for 60 seconds (designation “). An angle of 90° (Fig. 2) is called right; an angle less than 90° (Fig. 3) is called acute; an angle greater than 90° (Fig. 4) is called obtuse.

Straight lines forming a right angle are called mutually perpendicular. If the lines AB and MK are perpendicular, then this is denoted: AB MK.

Constructing an angle equal to a given one.

Before starting construction or solving any problem, regardless of the subject, you need to carry out analysis. Understand what the assignment says, read it thoughtfully and slowly. If after the first time you have doubts or something was not clear or clear but not completely, it is recommended to read it again. If you are doing an assignment in class, you can ask the teacher. Otherwise, your task, which you misunderstood, may not be solved correctly, or you may find something that is not what was required of you, and it will be considered incorrect and you will have to redo it. As for me - It’s better to spend a little more time studying the task than to redo the task all over again.

Analysis.

Let a be the given ray with vertex A, and the angle (ab) be the desired one. Let's choose points B and C on rays a and b, respectively. By connecting points B and C, we get triangle ABC. IN equal triangles the corresponding angles are equal, and hence the method of construction follows. If on the sides given angle in some convenient way select points C and B, from a given ray into a given half-plane construct a triangle AB 1 C 1 equal to ABC (and this can be done if you know all the sides of the triangle), then the problem will be solved.


When carrying out any constructions Be extremely careful and try to carry out all constructions carefully. Since any inconsistencies can result in some kind of errors, deviations, which can lead to an incorrect answer. And if the task of this type is performed for the first time, the error will be very difficult to find and fix.

Construction example first.

Let's draw a circle with its center at the vertex of this angle. Let B and C be the points of intersection of the circle with the sides of the angle. With radius AB we draw a circle with the center at point A 1 – the starting point of this ray. Let us denote the point of intersection of this circle with this ray as B 1 . Let us describe a circle with center at B 1 and radius BC. The intersection point C 1 of the constructed circles in the indicated half-plane lies on the side of the desired angle.


Triangles ABC and A 1 B 1 C 1 are equal on three sides. Angles A and A 1 are the corresponding angles of these triangles. Therefore, ∠CAB = ∠C 1 A 1 B 1

For greater clarity, you can consider the same constructions in more detail.

Construction example two.

The task remains to also set aside an angle from a given half-line into a given half-plane equal to a given angle.


Construction.

Step 1. Let's draw a circle with an arbitrary radius and centers at vertex A of a given angle. Let B and C be the points of intersection of the circle with the sides of the angle. And let's draw segment BC.


Step 2. Let's draw a circle of radius AB with the center at point O - the starting point of this half-line. Let us denote the point of intersection of the circle with the ray as B 1 .


Step 3. Now we describe a circle with center B 1 and radius BC. Let point C 1 be the intersection of the constructed circles in the indicated half-plane.

Step 4. Let's draw a ray from point O through point C 1. Angle C 1 OB 1 will be the desired one.


Proof.

Triangles ABC and OB 1 C 1 are congruent triangles with corresponding sides. And therefore angles CAB and C 1 OB 1 are equal.


Interesting fact:

In numbers.


In objects of the surrounding world, you first of all notice their individual properties that distinguish one object from another.

The abundance of particular, individual properties obscures the general properties inherent in absolutely all objects, and therefore it is always more difficult to detect such properties.

One of the most important general properties of objects is that all objects can be counted and measured. We reflect this general property objects in the concept of number.

People mastered the process of counting, that is, the concept of number, very slowly, over centuries, in a persistent struggle for their existence.

In order to count, one must not only have objects that can be counted, but also already have the ability to abstract when considering these objects from all their other properties except number, and this ability is the result of a long historical development based on experience.

Every person now learns to count with the help of numbers imperceptibly in childhood, almost simultaneously with the time he begins to speak, but this counting, which is familiar to us, has gone through a long path of development and has taken different forms.

There was a time when only two numerals were used to count objects: one and two. In the process of further expansion of the number system, parts were involved human body and first of all, fingers, and if this kind of “numbers” was not enough, then also sticks, stones and other things.

N. N. Miklouho-Maclay in his book "Trips" talks about a funny method of counting used by the natives of New Guinea:

Questions:

  1. Define angle?
  2. What types of angles are there?
  3. What is the difference between diameter and radius?

List of sources used:

  1. Mazur K. I. “Solving the main competition problems in mathematics of the collection edited by M. I. Skanavi”
  2. Mathematical savvy. B.A. Kordemsky. Moscow.
  3. L. S. Atanasyan, V. F. Butuzov, S. B. Kadomtsev, E. G. Poznyak, I. I. Yudina “Geometry, 7 – 9: textbook for educational institutions”

Worked on the lesson:

Levchenko V.S.

Poturnak S.A.

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Subjects > Mathematics > Mathematics 7th grade

In construction tasks we will consider the construction of a geometric figure, which can be done using a ruler and compass.

Using a ruler you can:

    arbitrary straight line;

    an arbitrary straight line passing through a given point;

    a straight line passing through two given points.

Using a compass, you can describe a circle of a given radius from a given center.

Using a compass you can plot a segment on a given line from a given point.

Let's consider the main construction tasks.

Task 1. Construct a triangle with given sides a, b, c (Fig. 1).

Solution. Using a ruler, draw an arbitrary straight line and take an arbitrary point B on it. Using a compass opening equal to a, we describe a circle with center B and radius a. Let C be the point of its intersection with the line. With a compass opening equal to c, we describe a circle from center B, and with a compass opening equal to b, we describe a circle from center C. Let A be the intersection point of these circles. Triangle ABC has sides equal to a, b, c.

Comment. In order for three straight line segments to serve as sides of a triangle, it is necessary that the largest of them be less than the amount the other two (and< b + с).

Task 2.

Solution. This angle with vertex A and the ray OM are shown in Figure 2.

Let's draw an arbitrary circle with its center at vertex A of the given angle. Let B and C be the points of intersection of the circle with the sides of the angle (Fig. 3, a). With radius AB we draw a circle with the center at point O - the starting point of this ray (Fig. 3, b). Let us denote the point of intersection of this circle with this ray as C 1 . Let us describe a circle with center C 1 and radius BC. Point B 1 of the intersection of two circles lies on the side of the desired angle. This follows from the equality Δ ABC = Δ OB 1 C 1 (the third sign of equality of triangles).

Task 3. Construct the bisector of this angle (Fig. 4).

Solution. From vertex A of a given angle, as from the center, we draw a circle of arbitrary radius. Let B and C be the points of its intersection with the sides of the angle. From points B and C we describe circles with the same radius. Let D be their intersection point, different from A. Ray AD bisects angle A. This follows from the equality Δ ABD = Δ ACD (the third criterion for the equality of triangles).

Task 4. Draw a perpendicular bisector to this segment (Fig. 5).

Solution. Using an arbitrary but identical compass opening (larger than 1/2 AB), we describe two arcs with centers at points A and B, which will intersect each other at some points C and D. The straight line CD will be the desired perpendicular. Indeed, as can be seen from the construction, each of the points C and D is equally distant from A and B; therefore, these points must lie on the perpendicular bisector to segment AB.

Task 5. Divide this segment in half. It is solved in the same way as problem 4 (see Fig. 5).

Task 6. Through a given point draw a line perpendicular to the given line.

Solution. There are two possible cases:

1) a given point O lies on a given straight line a (Fig. 6).

From point O we draw a circle with an arbitrary radius intersecting straight line a at points A and B. From points A and B we draw circles with the same radius. Let O 1 be the point of their intersection, different from O. We obtain OO 1 ⊥ AB. In fact, points O and O 1 are equidistant from the ends of the segment AB and, therefore, lie on the perpendicular bisector to this segment.

Often it is necessary to draw (“construct”) an angle that would be equal to a given angle, and the construction must be done without the help of a protractor, but using only a compass and a ruler. Knowing how to construct a triangle on three sides, we can solve this problem. Let it be on a straight line MN(Fig. 60 and 61) it is required to build at the point K angle equal to angle B. This means that it is necessary from the point K draw a straight line with a component MN angle equal to B.

To do this, mark a point on each side of a given angle, for example A And WITH, and connect A And WITH straight line. We get a triangle ABC. Let us now construct on a straight line MN this triangle so that its vertex IN was at the point TO: then at this point an angle will be constructed equal to the angle IN. Construct a triangle using three sides VS, VA And AC we know how: we postpone (Fig. 62) from the point TO segment KL, equal Sun; we get a point L; around K, as near the center, we describe a circle with a radius VA, and around L – radius SA. Full stop R we connect the intersections of the circles with TO and Z, we get a triangle KPL, equal to a triangle ABC; there's a corner in it TO= ug. IN.

This construction is performed faster and more conveniently if from the top IN lay down equal segments (with one dissolution of the compass) and, without moving its legs, describe a circle around the point with the same radius TO, like near the center.

How to split a corner in half

Suppose we need to divide an angle A(Fig. 63) into two equal parts using a compass and ruler, without using a protractor. We'll show you how to do it.

From the top A put equal segments on the sides of the angle AB And AC(Diagram 64; this is done by simply dissolving the compass). Then we place the tip of the compass at the points IN And WITH and describe arcs of equal radii intersecting at the point D. Straight connecting A and D divides the angle A in half.

Let's explain why this is. If the point D connect with IN and C (Fig. 65), then you get two triangles ADC And ADB, y which have a common side AD; side AB equal to side AC, A ВD equal to CD. The triangles are equal on three sides, which means the angles are equal. BAD And DAC, lying against equal sides ВD And CD. Therefore, straight AD divides the angle YOU in half.

Applications

12. Construct a 45° angle without a protractor. At 22°30’. At 67°30'.

Solution: Dividing the right angle in half, we get an angle of 45°. Dividing the 45° angle in half, we get an angle of 22°30’. By constructing the sum of the angles 45° + 22°30’, we get an angle of 67°30’.

How to construct a triangle using two sides and the angle between them

Suppose you need to find out on the ground the distance between two milestones A And IN(Devil 66), separated by an impassable swamp.

How to do this?

We can do this: choose a point away from the swamp WITH, from where both milestones are visible and distances can be measured AC And Sun. Corner WITH we measure using a special goniometric device (called a str o l b i e). According to these data, i.e., according to the measured sides A.C. And Sun and corner WITH between them, let's build a triangle ABC somewhere on convenient terrain as follows. Having measured one known side in a straight line (Fig. 67), for example AC, build with it at the point WITH corner WITH; on the other side of this angle the known side is measured Sun. The ends of the known sides, i.e. points A And IN connected by a straight line. The result is a triangle in which two sides and the angle between them have the dimensions specified in advance.

From the method of construction it is clear that only one triangle can be constructed using two sides and the angle between them. therefore, if two sides of one triangle are equal to two sides of another and the angles between these sides are the same, then such triangles can be superimposed on each other by all points, i.e. their third sides and other angles must also be equal. This means that the equality of two sides of triangles and the angle between them can serve as a sign of complete equality of these triangles. In short:

Triangles are equal on both sides and at the angle between them.

To construct any drawing or perform planar markings of a workpiece before processing it, it is necessary to carry out a number of graphic operations - geometric constructions.

In Fig. Figure 2.1 shows a flat part - a plate. To draw its drawing or mark a contour on a steel strip for subsequent manufacturing, you need to do it on the construction plane, the main ones are numbered with numbers written on the pointer arrows. In numbers 1 the construction of mutually perpendicular lines is indicated, which must be performed in several places, with the number 2 – drawing parallel lines, in numbers 3 – pairing these parallel lines with an arc of a certain radius, a number 4 – conjugation of an arc and a straight arc of a given radius, which in in this case equal to 10 mm, number 5 – pairing of two arcs with an arc of a certain radius.

As a result of performing these and other geometric constructions, the outline of the part will be drawn.

Geometric construction is a method of solving a problem in which the answer is obtained graphically without any calculations. Constructions are carried out with drawing (or marking) tools as carefully as possible, because the accuracy of the solution depends on this.

The lines specified by the conditions of the problem, as well as the constructions, are made solid thin, and the results of the construction are solid main ones.

When starting to make a drawing or marking, you must first determine which of the geometric constructions need to be applied in this case, i.e. analyze the graphic composition of the image.

Rice. 2.1.

Analysis of the graphic composition of the image called the process of dividing the execution of a drawing into separate graphic operations.

Identifying the operations required to construct a drawing makes it easier to choose how to execute it. If you need to draw, for example, the plate shown in Fig. 2.1, then analysis of the contour of its image leads us to the conclusion that we must apply the following geometric constructions: in five cases, draw mutually perpendicular center lines (figure 1 in a circle), in four cases draw parallel lines (number 2 ), draw two concentric circles (0 50 and 70 mm), in six cases construct mates of two parallel straight lines with arcs of a given radius (figure 3 ), and in four - the pairing of an arc and a straight arc of radius 10 mm (figure 4 ), in four cases, construct a pairing of two arcs with an arc of radius 5 mm (number 5 in a circle).

To carry out these constructions, you need to remember or repeat from the textbook the rules for drawing them.

In this case, it is advisable to choose a rational way to complete the drawing. Choosing a rational way to solve a problem reduces the time spent on work. For example, when constructing an equilateral triangle inscribed in a circle, a more rational method is to construct it using a crossbar and a square with an angle of 60° without first determining the vertices of the triangle (see Fig. 2.2, a, b). A less rational way to solve the same problem is using a compass and a crossbar with preliminary determination of the vertices of the triangle (see Fig. 2.2, V).

Dividing segments and constructing angles

Constructing right angles

It is rational to construct a 90° angle using a crossbar and a square (Fig. 2.2). To do this, it is enough to draw a straight line and restore a perpendicular to it using a square (Fig. 2.2, A). It is rational to build a perpendicular to the inclined segment by moving (Fig. 2.2, b) or turning (Fig. 2.2, V) square.

Rice. 2.2.

Construction of obtuse and acute angles

Rational methods for constructing angles of 120, 30 and 150, 60 and 120, 15 and 165, 75 and 105.45 and 135° are shown in Fig. 2.3, which shows the positions of the squares for constructing these angles.

Rice. 2.3.

Dividing an angle into two equal parts

From the vertex of the corner, describe an arc of a circle of arbitrary radius (Fig. 2.4).

Rice. 2.4.

From points ΜηΝ intersection of an arc with the sides of an angle with a compass solution larger than half the arc ΜΝ, make two intersecting at a point A serifs.

Through the received point A and the vertex of the angle draw a straight line (the bisector of the angle).

Dividing a right angle into three equal parts

From the top right angle describe an arc of a circle of arbitrary radius (Fig. 2.5). Without changing the angle of the compass, make notches from the points of intersection of the arc with the sides of the angle. Through the received points M And Ν and the vertex of the angle are drawn by straight lines.

Rice. 2.5.

In this way, only right angles can be divided into three equal parts.

Constructing an angle equal to a given one. From the top ABOUT given angle draw an arc of arbitrary radius R, intersecting the sides of the angle at points M And N(Fig. 2.6, A). Then draw a straight segment, which will serve as one of the sides of the new angle. From point ABOUT 1 on this straight line with the same radius R draw an arc, getting a point Ν 1 (Fig. 2.6, b). From this point describe an arc of radius R 1, equal to the chord MN. The intersection of arcs gives a point Μ 1, which is connected by a straight line to the vertex of the new angle (Fig. 2.6, b).

Rice. 2.6.

Dividing a line segment into two equal parts. Arcs are drawn from the ends of a given segment with a compass opening greater than half its length (Fig. 2.7). Straight line connecting the obtained points M And Ν, divides a segment into two equal parts and is perpendicular to it.

Rice. 2.7.

Constructing a perpendicular at the end of a straight line segment. From an arbitrary point O taken above the segment AB, describe a circle passing through a point A(end of a line segment) and intersecting the line at the point M(Fig. 2.8).

Rice. 2.8.

Through the received point M and center ABOUT circles draw a straight line until they meet the opposite side of the circle at a point N. Full stop N connect a straight line to a point A.

Dividing a line segment into any number of equal parts. From any end of a segment, for example from a point A, draw a straight line at an acute angle to it. On it, with a measuring compass, they lay down the right number equal segments of arbitrary size (Fig. 2.9). The last point is connected to the second end of the given segment (to the point IN). From all division points, using a ruler and a square, draw straight lines parallel to the straight line 9V, which will divide the segment AB into a given number of equal parts.

Rice. 2.9.

In Fig. Figure 2.10 shows how to apply this construction to mark the centers of holes evenly spaced on a straight line.

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