How to construct an angle using a compass. How to construct 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 second.

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, an angle is called an 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 second.

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 it is therefore 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 us 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.

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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 angles between them.

This - oldest geometric problem.

Step by step instructions

1st method. - Using the “golden” or “Egyptian” triangle. The sides of this triangle have the aspect ratio 3:4:5, and the angle is strictly 90 degrees. This quality was widely used by the ancient Egyptians and other ancient cultures.

Ill.1. Construction of the Golden or Egyptian Triangle

• We manufacture three measurements (or rope compasses - a rope on two nails or pegs) with lengths 3; 4; 5 meters. The ancients often used the method of tying knots with equal distances between them as units of measurement. Unit of length - " nodule».
• We drive a peg at point O and attach the measure “R3 - 3 knots” to it.
• We stretch the rope along the known boundary - towards the proposed point A.
• At the moment of tension on the border line - point A, we drive in a peg.
• Then - again from point O, stretch the measure R4 - along the second border. We don’t drive the peg in yet.
• After this, we stretch the measure R5 - from A to B.
• We drive a peg at the intersection of measurements R2 and R3. – This is the desired point B – third vertex of the golden triangle, with sides 3;4;5 and with a right angle at point O.

2nd method. Using a compass.

The compass may be rope or pedometer. Cm:

Our compass pedometer has a step of 1 meter.

Ill.2. Compass pedometer

Construction - also according to Ill. 1.

• From the reference point - point O - the neighbor's corner, draw a segment of arbitrary length - but larger than the radius of the compass = 1m - in each direction from the center (segment AB).
• We place the leg of the compass at point O.
• We draw a circle with radius (compass pitch) = 1 m. It is enough to draw short arcs - 10-20 centimeters each, at the intersection with the marked segment (through points A and B). With this action we found equidistant points from the center- A and B. The distance from the center does not matter here. You can simply mark these points with a tape measure.
• Next, you need to draw arcs with centers at points A and B, but with a slightly (arbitrarily) larger radius than R=1m. You can reconfigure our compass to a larger radius if it has an adjustable pitch. But for such a small current task, I wouldn’t want to “pull” it. Or when there is no adjustment. Can be done in half a minute rope compass.
• We place the first nail (or the leg of a compass with a radius greater than 1 m) alternately at points A and B. And draw two arcs with the second nail - in a taut state of the rope - so that they intersect with each other. It is possible at two points: C and D, but one is enough - C. And again, short serifs at the intersection at point C will suffice.
• Draw a straight line (segment) through points C and D.
• All! The resulting segment, or straight line, is exact direction north:). Sorry, - at a right angle.
• The figure shows two cases of boundary discrepancy across a neighbor's property. Ill. 3a shows a case where a neighbor’s fence moves away from the desired direction to its detriment. On 3b - he climbed onto your site. In situation 3a, it is possible to construct two “guide” points: both C and D. In situation 3b, only C.
• Place a peg at corner O, and a temporary peg at point C, and stretch a cord from C to the rear boundary of the site. – So that the cord barely touches peg O. By measuring from point O - in direction D, the length of the side according to the general plan, you will get a reliable rear right corner of the site.

Ill.3. Construction right angle– from the neighbor’s corner, using a pedometer and a rope compass

If you have a compass-pedometer, then you can do without rope altogether. In the previous example, we used the rope one to draw arcs of a larger radius than those of the pedometer. More because these arcs must intersect somewhere. In order for the arcs to be drawn with a pedometer with the same radius - 1m with a guarantee of their intersection, it is necessary that points A and B are inside the circle with R = 1m.

• Then measure these equidistant points roulette- V different sides from the center, but always along line AB (neighbor’s fence line). The closer points A and B are to the center, the farther the guide points C and D are from it, and the more accurate the measurements. In the figure, this distance is taken to be about a quarter of the pedometer radius = 260mm.

Ill.4. Constructing a right angle using a compass-pedometer and tape measure

• This scheme of actions is no less relevant when constructing any rectangle, in particular the contour of a rectangular foundation. You will receive it perfect. Its diagonals, of course, need to be checked, but isn't the effort reduced? – Compared to when the diagonals, corners and sides of the foundation contour are moved back and forth until the corners meet..

Actually, we solved a geometric problem on earth. To make your actions more confident on the site, practice on paper - using a regular compass. Which is basically no different.