Formula for finding an inscribed angle. Circle. Central and inscribed angle

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

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

In other words, inscribed angle includes as many angular degrees, minutes and seconds as arc degrees, minutes and seconds are contained in half the arc on which it rests. To justify this, let us analyze three cases:

First case:

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

Second case:

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

Third case:

Center O is located outside inscribed angle ABC. Drawing the diameter BD, we will have:∠ABC = ∠ABD - ∠CBD . But angles ABD and CBD are measured based on the previously justified half arc AD and CD. And since ∠ABC is measured by (AD-CD)/2, that is, half the arc AC.

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

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

Today we will look at another type of problems 6 - this time with a circle. Many students do not like them and find them difficult. And completely in vain, since such problems are solved elementary, if you know some theorems. Or they don’t dare at all if you don’t know them.

Before talking about the main properties, let me remind you of the definition:

An inscribed angle is one whose vertex lies on the circle itself, and whose sides cut out a chord on this circle.

A central angle is any angle with its vertex at the center of the circle. Its sides also intersect this circle and carve a chord on it.

So, the concepts of inscribed and central angles are inextricably linked with the circle and the chords inside it. And now the main statement:

Theorem. The central angle is always twice the inscribed angle, based on the same arc.

Despite the simplicity of the statement, there is a whole class of problems 6 that can be solved using it - and nothing else.

Task. Find an acute inscribed angle subtended by a chord equal to the radius of the circle.

Let AB be the chord under consideration, O the center of the circle. Additional construction: OA and OB are the radii of the circle. We get:

Consider triangle ABO. In it AB = OA = OB - all sides are equal to the radius of the circle. Therefore, triangle ABO is equilateral, and all angles in it are 60°.

Let M be the vertex of the inscribed angle. Since angles O and M rest on the same arc AB, the inscribed angle M is 2 times smaller than the central angle O. We have:

M = O: 2 = 60: 2 = 30

Task. The central angle is 36° greater than the inscribed angle subtended by the same arc of a circle. Find the inscribed angle.

Let us introduce the following notation:

1. AB is the chord of the circle;
2. Point O is the center of the circle, so angle AOB is the central angle;
3. Point C is the vertex of the inscribed angle ACB.

Since we are looking for the inscribed angle ACB, let's denote it ACB = x. Then the central angle AOB is x + 36. On the other hand, the central angle is 2 times the inscribed angle. We have:

AOB = 2 · ACB ;
x + 36 = 2 x ;
x = 36.

So we found the inscribed angle AOB - it is equal to 36°.

A circle is an angle of 360°

Having read the subtitle, knowledgeable readers will probably now say: “Ugh!” Indeed, comparing a circle with an angle is not entirely correct. To understand what we're talking about, take a look at the classic trigonometric circle:

What is this picture for? And besides, a full rotation is an angle of 360 degrees. And if you divide it, say, into 20 equal parts, then the size of each of them will be 360: 20 = 18 degrees. This is exactly what is required to solve problem B8.

Points A, B and C lie on the circle and divide it into three arcs, the degree measures of which are in the ratio 1: 3: 5. Find the greater angle of triangle ABC.

First, let's find the degree measure of each arc. Let the smaller one be x. In the figure this arc is designated AB. Then the remaining arcs - BC and AC - can be expressed in terms of AB: arc BC = 3x; AC = 5x. In total, these arcs give 360 ​​degrees:

AB + BC + AC = 360;
x + 3x + 5x = 360;
9x = 360;
x = 40.

Now consider a large arc AC that does not contain point B. This arc, like the corresponding central angle AOC, is 5x = 5 40 = 200 degrees.

Angle ABC is the largest of all angles in a triangle. It is an inscribed angle subtended by the same arc as the central angle AOC. This means that angle ABC is 2 times less than AOC. We have:

ABC = AOC: 2 = 200: 2 = 100

This will be the degree measure larger angle in triangle ABC.

Circle circumscribed around a right triangle

Many people forget this theorem. But in vain, because some B8 problems cannot be solved at all without it. More precisely, they are solved, but with such a volume of calculations that you would rather fall asleep than reach the answer.

Theorem. Center of the circumscribed circle right triangle, lies in the middle of the hypotenuse.

What follows from this theorem?

1. The midpoint of the hypotenuse is equidistant from all the vertices of the triangle. This is a direct consequence of the theorem;
2. The median drawn to the hypotenuse divides the original triangle into two isosceles triangles. This is exactly what is required to solve problem B8.

In triangle ABC we draw the median CD. Angle C is 90° and angle B is 60°. Find angle ACD.

Since angle C is 90°, triangle ABC is a right triangle. It turns out that CD is the median drawn to the hypotenuse. This means that triangles ADC and BDC are isosceles.

In particular, consider triangle ADC. In it AD = CD. But in an isosceles triangle, the angles at the base are equal - see “Problem B8: Line segments and angles in triangles.” Therefore, the desired angle ACD = A.

So, it remains to find out what the angle A is equal to. To do this, let's turn again to the original triangle ABC. Let's denote the angle A = x. Since the sum of the angles in any triangle is 180°, we have:

A + B + BCA = 180;
x + 60 + 90 = 180;
x = 30.

Of course, the last problem can be solved differently. For example, it is easy to prove that triangle BCD is not just isosceles, but equilateral. So angle BCD is 60 degrees. Hence angle ACD is 90 − 60 = 30 degrees. As you can see, you can use different isosceles triangles, but the answer will always be the same.

Most often, the process of preparing for the Unified State Exam in mathematics begins with a repetition of basic definitions, formulas and theorems, including on the topic “Central and inscribed angles in a circle.” As a rule, this section of planimetry is studied in high school. It is not surprising that many students face the need to repeat basic concepts and theorems on the topic “Central angle of a circle.” Having understood the algorithm for solving such problems, schoolchildren will be able to count on receiving competitive scores based on the results of passing the unified state exam.

How to easily and effectively prepare for passing the certification test?

When 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 school textbook available at hand. And searching for formulas on the Internet sometimes takes a lot of time.

Our team will help you “pump up” your skills and improve your knowledge in such a difficult section of geometry as planimetry educational portal. “Shkolkovo” offers high school students and their teachers a new way to build the process of preparing for the unified state exam. All basic material is presented by our specialists in the most accessible form. After reading the information in the “Theoretical Background” 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 practice skills, we recommend performing appropriate exercises. Large selection tasks for finding the value of an angle inscribed in a circle and other parameters are presented in the “Catalog” section. For each exercise, our experts wrote out a detailed 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 Unified State Exam by practicing exercises, for example, to find the magnitude of a central angle and the length of an arc of a circle, online, from 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.

The concept of inscribed and central angle

Let us first introduce the concept of a central angle.

Note 1

Note that the degree measure of a central angle is equal to the degree measure of the arc on which it rests.

Let us now introduce the concept of an inscribed angle.

Definition 2

An angle whose vertex lies on a circle and whose sides intersect the same circle is called an inscribed angle (Fig. 2).

Figure 2. Inscribed angle

Inscribed angle theorem

Theorem 1

The degree measure of an inscribed angle is equal to half the degree measure of the arc on which it rests.

Proof.

Let us be given a circle with center at point $O$. Let's denote the inscribed angle $ACB$ (Fig. 2). The following three cases are possible:

• Ray $CO$ coincides with any side of the angle. Let this be the side $CB$ (Fig. 3).

Figure 3.

In this case, the arc $AB$ is less than $(180)^(()^\circ )$, therefore the central angle $AOB$ is equal to the arc $AB$. Since $AO=OC=r$, then the triangle $AOC$ is isosceles. This means that the base angles $CAO$ and $ACO$ are equal to each other. According to the theorem on the external angle of a triangle, we have:

• Beam $CO$ divides internal corner at two angles. Let it intersect the circle at point $D$ (Fig. 4).

Figure 4.

We get

• Ray $CO$ does not divide the interior angle into two angles and does not coincide with any of its sides (Fig. 5).

Figure 5.

Let us consider angles $ACD$ and $DCB$ separately. According to what was proved in point 1, we get

We get

The theorem has been proven.

Let's give consequences from this theorem.

Corollary 1: Inscribed angles that rest on the same arc are equal to each other.

Corollary 2: An inscribed angle that subtends a diameter is a right angle.

Instructions

If the radius (R) of the circle and the length of the arc (L) corresponding to the desired central angle (θ) are known, it can be calculated both in degrees and in radians. The total is determined by the formula 2*π*R and corresponds to a central angle of 360° or two Pi numbers, if radians are used instead of degrees. Therefore, proceed from the proportion 2*π*R/L = 360°/θ = 2*π/θ. Express from it the central angle in radians θ = 2*π/(2*π*R/L) = L/R or degrees θ = 360°/(2*π*R/L) = 180*L/(π* R) and calculate using the resulting formula.

Based on the length of the chord (m) connecting the points that determine the central angle (θ), its value can also be calculated if the radius (R) of the circle is known. To do this, consider a triangle formed by two radii and . This is an isosceles triangle, everyone is known, but you need to find the angle opposite the base. The sine of its half is equal to the ratio of the length of the base - the chord - to twice the length of the side - the radius. Therefore, use the inverse sine function for calculations - arcsine: θ = 2*arcsin(½*m/R).

The central angle can be specified in fractions of a revolution or from a rotated angle. For example, if you need to find the central angle corresponding to a quarter of a full revolution, divide 360° by four: θ = 360°/4 = 90°. The same value in radians should be 2*π/4 ≈ 3.14/2 ≈ 1.57. The unfolded angle is equal to half a full revolution, therefore, for example, the central angle corresponding to a quarter of it will be half the values ​​​​calculated above in both degrees and radians.

The inverse of sine is called a trigonometric function arcsine. It can take values ​​within half the number Pi, both positive and negative. negative side when measured in radians. When measured in degrees, these values ​​will be respectively in the range from -90° to +90°.

Instructions

Some “round” values ​​do not need to be calculated; they are easier to remember. For example:- if the function argument equal to zero, then the value of the arcsine of it is also zero; - from 1/2 is equal to 30° or 1/6 Pi, if measured; - the arcsine of -1/2 is equal to -30° or -1/6 of the number Pi in; - arcsine from 1 is equal to 90° or 1/2 of Pi in radians; - arcsine of -1 is equal to -90° or -1/2 of Pi in radians;

To measure the values ​​of this function from other arguments, the easiest way is to use a standard Windows calculator, if you have one at hand. To start, open the main menu on the “Start” button (or by pressing the WIN key), go to the “All Programs” section, and then to the “Accessories” subsection and click “Calculator”.

Switch the calculator interface to the operating mode that allows you to calculate trigonometric functions. To do this, open the “View” section in its menu and select “Engineering” or “Scientific” (depending on the type of operating system).

Enter the value of the argument from which the arctangent should be calculated. This can be done by clicking the buttons on the calculator interface with the mouse, or by pressing the keys on , or by copying the value (CTRL + C) and then pasting it (CTRL + V) into the input field of the calculator.

Select the units of measurement in which you need to obtain the result of the function calculation. Below the input field there are three options, from which you need to select (by clicking it with the mouse) one - , radians or rads.

Check the checkbox that inverts the functions indicated on the calculator interface buttons. Next to it is a short inscription Inv.

Click the sin button. The calculator will invert the function associated with it, perform the calculation and present you with the result in the specified units.

Video on the topic

One of the common geometric problems is calculating the area of ​​a circular segment - the part of the circle bounded by a chord and the corresponding chord by an arc of a circle.

The area of ​​a circular segment is equal to the difference between the area of ​​the corresponding circular sector and the area of ​​the triangle formed by the radii of the sector corresponding to the segment and the chord limiting the segment.

Example 1

The length of the chord subtending the circle is equal to the value a. The degree measure of the arc corresponding to the chord is 60°. Find the area of ​​the circular segment.

Solution

A triangle formed by two radii and a chord is isosceles, so the height drawn from the vertex of the central angle to the side of the triangle is formed by a chord, will also be the bisector of the central angle, dividing it in half, and the median, dividing the chord in half. Knowing that the sine of the angle is equal to the ratio of the opposite leg to the hypotenuse, we can calculate the radius:

Sin 30°= a/2:R = 1/2;

Sc = πR²/360°*60° = πa²/6

S▲=1/2*ah, where h is the height drawn from the vertex of the central angle to the chord. According to the Pythagorean theorem h=√(R²-a²/4)= √3*a/2.

Accordingly, S▲=√3/4*a².

The area of ​​the segment, calculated as Sreg = Sc - S▲, is equal to:

Sreg = πa²/6 - √3/4*a²

Substituting numeric value Instead of the value a, you can easily calculate the numerical value of the segment area.

Example 2

The radius of the circle is equal to a. The degree measure of the arc corresponding to the segment is 60°. Find the area of ​​the circular segment.

Solution:

The area of ​​the sector corresponding to a given angle can be calculated using the following formula:

Sc = πа²/360°*60° = πa²/6,

The area of ​​the triangle corresponding to the sector is calculated as follows:

S▲=1/2*ah, where h is the height drawn from the vertex of the central angle to the chord. According to the Pythagorean theorem h=√(a²-a²/4)= √3*a/2.

Accordingly, S▲=√3/4*a².

And finally, the area of ​​the segment, calculated as Sreg = Sc - S▲, is equal to:

Sreg = πa²/6 - √3/4*a².

The solutions in both cases are almost identical. Thus, we can conclude that to calculate the area of ​​a segment in the simplest case, it is enough to know the value of the angle corresponding to the arc of the segment and one of two parameters - either the radius of the circle or the length of the chord subtending the arc of the circle forming the segment.

Sources:

• Segment - geometry