What does the central corner look like? Circle

Angle ABC is an inscribed angle. It rests on the arc AC, enclosed between its sides (Fig. 330).

Theorem. An inscribed angle is measured by half the arc it intercepts.

This should be understood as follows: an inscribed angle contains as many angular degrees, minutes and seconds as arc degrees, minutes and seconds are contained in the half of the arc on which it rests.

In proving this theorem, we need to consider three cases.

First case. The center of the circle lies on the side of the inscribed angle (Fig. 331).

Let ∠ABC be an inscribed angle and the center of circle O lies on side BC. It is required to prove that it is measured by half the arc AC.

Connect point A to the center of the circle. We get the isosceles \(\Delta\)AOB, in which AO = OB, as the radii of the same circle. Therefore, ∠A = ∠B.

∠AOC is external to triangle AOB, so ∠AOC = ∠A + ∠B, and since angles A and B are equal, ∠B is 1/2 ∠AOC.

But ∠AOC is measured by arc AC, therefore ∠B is measured by half of arc AC.

For example, if \(\breve(AC)\) contains 60°18', then ∠B contains 30°9'.

Second case. The center of the circle lies between the sides of the inscribed angle (Fig. 332).

Let ∠ABD be an inscribed angle. The center of circle O lies between its sides. It is required to prove that ∠ABD is measured by half of the arc AD.

To prove this, let's draw the diameter BC. Angle ABD split into two angles: ∠1 and ∠2.

∠1 is measured by half of the arc AC, and ∠2 is measured by half of the arc CD, therefore, the entire ∠ABD is measured by 1/2 \(\breve(AC)\) + 1/2 \(\breve(CD)\), i.e. half of the arc AD.

For example, if \(\breve(AD)\) contains 124°, then ∠B contains 62°.

Third case. The center of the circle lies outside the inscribed angle (Fig. 333).

Let ∠MAD be an inscribed angle. The center of circle O is outside the corner. It is required to prove that ∠MAD is measured by half of the arc MD.

To prove this, let us draw the diameter AB. ∠MAD = ∠MAB - ∠DAB. But ∠MAB measures 1/2 \(\breve(MB)\) and ∠DAB measures 1/2 \(\breve(DB)\).

Therefore, ∠MAD measures 1 / 2 (\(\breve(MB) - \breve(DB))\), i.e. 1 / 2 \(\breve(MD)\).

For example, if \(\breve(MD)\) contains 48° 38", then ∠MAD contains 24° 19' 8".

Consequences
1. All inscribed angles based on the same arc are equal to each other, since they are measured by half of the same arc (Fig. 334, a).

2. An inscribed angle based on a diameter is a right angle because it is based on half a circle. Half of the circle contains 180 arc degrees, which means that the angle based on the diameter contains 90 angular degrees (Fig. 334, b).

Central corner is the angle whose vertex is at the center of the circle.
Inscribed angle An angle whose vertex lies on the circle and whose sides intersect it.

The figure shows central and inscribed angles, as well as their most important properties.

So, the value of the central angle is equal to the angular value of the arc on which it rests. This means that a central angle of 90 degrees will be based on an arc equal to 90 °, that is, a circle. The central angle, equal to 60°, is based on an arc of 60 degrees, that is, on the sixth part of the circle.

The value of the inscribed angle is two times less than the central one based on the same arc.

Also, to solve problems, we need the concept of "chord".

Equal central angles are supported by equal chords.

1. What is the inscribed angle based on the diameter of the circle? Give your answer in degrees.

An inscribed angle based on a diameter is a right angle.

2. The central angle is 36° greater than the acute inscribed angle based on the same circular arc. Find the inscribed angle. Give your answer in degrees.

Let the central angle be x, and the inscribed angle based on the same arc be y.

We know that x = 2y.
Hence 2y = 36 + y,
y = 36.

3. The radius of the circle is 1. Find the value of an obtuse inscribed angle based on a chord equal to . Give your answer in degrees.

Let the chord AB be . An obtuse inscribed angle based on this chord will be denoted by α.
In triangle AOB, sides AO and OB are equal to 1, side AB is equal to . We have seen such triangles before. Obviously, the triangle AOB is right-angled and isosceles, that is, the angle AOB is 90 °.
Then the arc ASV is equal to 90°, and the arc AKB is equal to 360° - 90° = 270°.
The inscribed angle α rests on the AKB arc and is equal to half the angular value of this arc, i.e. 135°.

Answer: 135.

4. The chord AB divides the circle into two parts, the degree values ​​of which are related as 5:7. At what angle is this chord visible from point C, which belongs to the smaller arc of the circle? Give your answer in degrees.

The main thing in this task is the correct drawing and understanding of the condition. How do you understand the question: “At what angle is the chord visible from point C?”
Imagine that you are sitting at point C and you need to see everything that happens on chord AB. So, as if the chord AB is a screen in a cinema :-)
Obviously, you need to find the angle ACB.
The sum of the two arcs into which the chord AB divides the circle is 360°, i.e.
5x + 7x = 360°
Hence x = 30°, and then the inscribed angle ACB rests on an arc equal to 210°.
The value of the inscribed angle is equal to half the angular value of the arc on which it rests, which means that the angle ACB is equal to 105°.

Most often, the process of preparing for the exam in mathematics begins with a repetition of the basic definitions, formulas and theorems, including the topic "Central and inscribed in a circle angle." As a rule, this section of planimetry is studied in high school. It is not surprising that many students are faced with the need to repeat the basic concepts and theorems on the topic "Central Angle of a Circle". Having figured out the algorithm for solving such problems, schoolchildren will be able to count on getting competitive points based on the results of passing the unified state exam.

How to easily and effectively prepare for the certification test?

Studying before passing the unified state exam, many high school students are faced with the problem of finding the necessary information on the topic "Central and inscribed angles in a circle." Not always a school textbook is at hand. And searching for formulas on the Internet sometimes takes a lot of time.

Our educational portal will help you to “pump” your skills and improve your knowledge in such a difficult section of geometry as planimetry. Shkolkovo invites high school students and their teachers to build the process of preparing for the unified state exam in a new way. All basic material is presented by our specialists in the most accessible form. After reviewing the information in the "Theoretical Reference" section, students will learn what properties the central angle of a circle has, how to find its value, etc.

Then, to consolidate the acquired knowledge and develop skills, we recommend that you perform the appropriate exercises. A large selection of tasks for finding the value of an angle inscribed in a circle and other parameters is presented in the Catalog section. For each exercise, our experts wrote down a detailed course of the solution and indicated the correct answer. The list of tasks on the site is constantly supplemented and updated.

High school students can prepare for the exam by practicing exercises, for example, finding the value of the central angle and the length of the arc of a circle, online, being in any Russian region.

If necessary, the completed task can be saved in the "Favorites" section in order to return to it later and once again analyze the principle of its solution.

This is the angle formed by two chords originating at one point on the circle. An inscribed angle is said to be relies on an arc enclosed between its sides.

Inscribed angle equal to half of the arc on which it rests.

In other words, inscribed angle includes as many degrees, minutes and seconds as arc degrees, minutes and seconds are enclosed in half of the arc on which it relies. For justification, we analyze three cases:

First case:

Center O is located on the side inscribed angle ABS. Drawing the radius AO, we get ΔABO, in which OA = OB (as radii) and, accordingly, ∠ABO = ∠BAO. In relation to this triangle, the angle AOC is external. And so, it is equal to the sum of the angles ABO and BAO, or equal to the double angle ABO. So ∠ABO is half central corner AOC. But this angle is measured by arc AC. That is, the inscribed angle ABC is measured by half the arc AC.

Second case:

The center O is located between the sides inscribed angle ABC. Having drawn the diameter BD, we will divide the angle ABC into two angles, of which, according to the established in the first case, one is measured by half arcs AD, and the other half of the arc CD. And accordingly, the angle ABC is measured by (AD + DC) / 2, i.e. 1/2 AC.

Third case:

Center O is located outside inscribed angle ABS. Having drawn the diameter BD, we will have: ∠ABС = ∠ABD - ∠CBD . But the angles ABD and CBD are measured, based on the previously substantiated halves arcs AD and CD. And since ∠ABС is measured by (AD-CD)/2, that is, half of the AC arc.

Consequence 1. Any , based on the same arc are the same, that is, they are equal to each other. Since each of them is measured by half of the same arcs .

Consequence 2. Inscribed angle, based on the diameter - right angle. Since each such angle is measured by half a semicircle and, accordingly, contains 90 °.

Inscribed angle, problem theory. Friends! In this article we will talk about tasks, for the solution of which it is necessary to know the properties of an inscribed angle. This is a whole group of tasks, they are included in the exam. Most of them are solved very simply, in one step.

There are more difficult tasks, but they will not present much difficulty for you, you need to know the properties of the inscribed angle. Gradually, we will analyze all the prototypes of tasks, I invite you to the blog!

Now the necessary theory. Recall what a central and inscribed angle, chord, arc, on which these angles rely:

The central angle in a circle is called a flat angle withpinnacle at its center.

The part of a circle that is inside a flat cornercalled an arc of a circle.

The degree measure of an arc of a circle is the degree measurecorresponding central angle.

An angle is called inscribed in a circle if the vertex of the angle lieson a circle, and the sides of the angle intersect this circle.

A line segment that connects two points on a circle is calledchord. The longest chord passes through the center of the circle and is calleddiameter.

To solve problems for angles inscribed in a circle,you need to know the following properties:

1. The inscribed angle is equal to half the central angle based on the same arc.

2. All inscribed angles based on the same arc are equal.

3. All inscribed angles based on the same chord, the vertices of which lie on the same side of this chord, are equal.

4. Any pair of angles based on the same chord, whose vertices lie on opposite sides of the chord, add up to 180 °.

Corollary: Opposite angles of a quadrilateral inscribed in a circle add up to 180 degrees.

5. All inscribed angles based on the diameter are straight.

In general, this property is a consequence of property (1), this is its special case. Look - the central angle is equal to 180 degrees (and this developed angle is nothing more than a diameter), which means that according to the first property, the inscribed angle C is equal to its half, that is, 90 degrees.

Knowledge of this property helps in solving many problems and often allows you to avoid unnecessary calculations. Having mastered it well, you will be able to solve more than half of this type of problems orally. Two consequences that can be made:

Corollary 1: if a triangle is inscribed in a circle and one of its sides coincides with the diameter of this circle, then the triangle is right-angled (the vertex of the right angle lies on the circle).

Corollary 2: The center of the circle circumscribed about a right triangle coincides with the midpoint of its hypotenuse.

Many prototypes of stereometric problems are also solved by using this property and these corollaries. Remember the fact itself: if the diameter of a circle is a side of an inscribed triangle, then this triangle is right-angled (the angle opposite the diameter is 90 degrees). You can draw all the other conclusions and consequences yourself, you do not need to teach them.

As a rule, half of the problems for an inscribed angle are given with a sketch, but without notation. To understand the process of reasoning when solving problems (below in the article), the designations of vertices (corners) are introduced. On the exam, you can not do this.Consider the tasks:

What is an acute inscribed angle that intercepts a chord equal to the radius of the circle? Give your answer in degrees.

Let's build a central angle for a given inscribed angle, denote the vertices:

According to the property of an angle inscribed in a circle:

The angle AOB is equal to 60 0, since the triangle AOB is equilateral, and in an equilateral triangle all angles are equal to 60 0 . The sides of the triangle are equal, since the condition says that the chord is equal to the radius.

Thus, the inscribed angle DIA is 30 0 .

Answer: 30

Find the chord on which the angle 30 0 rests, inscribed in a circle of radius 3.

This is essentially the inverse problem (of the previous one). Let's build a central corner.

It is twice as large as the inscribed one, that is, the angle AOB is 60 0 . From this we can conclude that the triangle AOB is equilateral. Thus, the chord is equal to the radius, that is, three.

Answer: 3

The radius of the circle is 1. Find the value of an obtuse inscribed angle based on a chord equal to the root of two. Give your answer in degrees.

Let's build the central angle:

Knowing the radius and chord, we can find the central angle DIA. This can be done using the law of cosines. Knowing the central angle, we can easily find the inscribed angle ACB.

Cosine theorem: the square of any side of a triangle is equal to the sum of the squares of the other two sides, without doubling the product of those sides times the cosine of the angle between them.

Therefore, the second central angle is 360 0 – 90 0 = 270 0 .

According to the property of an inscribed angle, the angle DIA is equal to its half, that is, 135 degrees.

Answer: 135

Find the chord on which the angle of 120 degrees, the root of three, is inscribed in a circle of radius.

Connect points A and B to the center of the circle. Let's call it O:

We know the radius and inscribed angle DIA. We can find the central angle AOB (greater than 180 degrees), then find the angle AOB in triangle AOB. And then, using the cosine theorem, calculate AB.

By the property of an inscribed angle, the central angle AOB (which is greater than 180 degrees) will be equal to twice the inscribed angle, that is, 240 degrees. This means that the angle AOB in the triangle AOB is 360 0 - 240 0 = 120 0 .

According to the law of cosines:

Answer:3

Find the inscribed angle based on the arc that is 20% of the circle. Give your answer in degrees.

By the property of an inscribed angle, it is half the size of the central angle based on the same arc, in this case we are talking about the arc AB.

It is said that the arc AB is 20 percent of the circumference. This means that the central angle AOB is also 20 percent of 360 0 .* A circle is a 360 degree angle. Means,

Thus, the inscribed angle ACB is 36 degrees.

Answer: 36

arc of a circle AC, not containing points B, is 200 degrees. And the arc of the circle BC, which does not contain points A, is 80 degrees. Find the inscribed angle ACB. Give your answer in degrees.

Let us denote for clarity the arcs whose angular measures are given. The arc corresponding to 200 degrees is blue, the arc corresponding to 80 degrees is red, the rest of the circle is yellow.

Thus, the degree measure of the arc AB (yellow), and hence the central angle AOB is: 360 0 – 200 0 – 80 0 = 80 0 .

The inscribed angle DAB is half the central angle AOB, that is, equal to 40 degrees.

Answer: 40

What is the inscribed angle based on the diameter of the circle? Give your answer in degrees.