Inscribed angle, theory of the problem. Friends! In this article we will talk about tasks for which you need to know the properties of an inscribed angle. This is a whole group of tasks, they are included in the Unified State Exam. Most of them can be solved very simply, in one action.
There are more difficult problems, but they won’t present much difficulty for you; you need to know the properties of an inscribed angle. Gradually we will analyze all the prototypes of tasks, I invite you to the blog!
Now the necessary theory. Let us remember what a central and inscribed angle, a chord, an arc are, on which these angles rest:
The central angle in a circle is a plane angle withapex at its center.
The part of a circle located inside a plane anglecalled an arc of a circle.
The degree measure of an arc of a circle is called the degree measurethe corresponding central angle.
An angle is said to be inscribed in a circle if the vertex of the angle lieson a circle, and the sides of the angle intersect this circle.
A segment connecting two points on a circle is calledchord. The largest chord passes through the center of the circle and is calleddiameter.
To solve problems involving 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 subtending the same arc are equal.
3. All inscribed angles based on the same chord and whose vertices lie on the same side of this chord are equal.
4. Any pair of angles based on the same chord, the vertices of which lie on opposite sides of the chord, add up to 180°.
Corollary: the opposite angles of a quadrilateral inscribed in a circle add up to 180 degrees.
5. All inscribed angles subtended by a diameter are right angles.
In general, this property is a consequence of property (1); this is its special case. Look - central angle is equal to 180 degrees (and this unfolded angle is nothing more than a diameter), which means, according to the first property, the inscribed angle C is equal to its half, that is, 90 degrees.
Knowing 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 the problems of this type orally. Two conclusions that can be drawn:
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 (vertex right angle lies on the circle).
Corollary 2: the center of the described about right triangle circle coincides with the middle of its hypotenuse.
Many prototypes of stereometric problems are also solved by using this property and these consequences. 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 other conclusions and consequences yourself; you don’t need to teach them.
As a rule, half of the problems on inscribed angles are given with a sketch, but without symbols. To understand the reasoning process when solving problems (below in the article), notations for vertices (angles) are introduced. You don't have to do this on the Unified State Examination.Let's consider the tasks:
What is the value of an acute inscribed angle subtended by a chord equal to the radius of the circle? Give your answer in degrees.
Let's construct a central angle for a given inscribed angle and designate the vertices:
According to the property of an angle inscribed in a circle:
Angle AOB is equal to 60 0, since the triangle AOB is equilateral, and in 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 ACB is equal to 30 0.
Answer: 30
Find the chord supported by an angle of 30 0 inscribed in a circle of radius 3.
This is essentially the inverse problem (of the previous one). Let's construct the central angle.
It is twice as large as the inscribed one, that is, angle AOB is equal to 60 0. From this we can conclude that 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 magnitude of the obtuse inscribed angle subtended by the chord equal to the root of two. Give your answer in degrees.
Let's construct the central angle:
Knowing the radius and chord, we can find the central angle ASV. This can be done using the cosine theorem. Knowing the central angle, we can easily find the inscribed angle ACB.
Cosine theorem: square any side of the triangle equal to the sum squares of the other two sides, without doubling the product of these sides by the cosine of the angle between them.
Therefore, the second central angle is 360 0 – 90 0 = 270 0 .
Angle ACB, by the property of an inscribed angle, is equal to half of it, that is, 135 degrees.
Answer: 135
Find the chord subtended by an angle of 120 degrees inscribed in a circle of radius root of three.
Let's connect points A and B to the center of the circle. Let's denote it as O:
We know the radius and inscribed angle ASV. 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.
According to 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 angle AOB in triangle AOB is equal to 360 0 – 240 0 = 120 0.
According to the cosine theorem:
Answer:3
Find the inscribed angle subtended by an arc that is 20% of the circle. Give your answer in degrees.
According to the property of the inscribed angle, it is half the size of the central angle based on the same arc, in in this case We are talking about arc AB.
It is said that 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 an angle of 360 degrees. Means,
Thus, the inscribed angle ACB is 36 degrees.
Answer: 36
Arc of a circle A.C., not containing a point B, is 200 degrees. And the arc of a circle BC, not containing a point A, is 80 degrees. Find the inscribed angle ACB. Give your answer in degrees.
For clarity, let us denote the arcs whose angular measures are given. Arc corresponding to 200 degrees – blue, the arc corresponding to 80 degrees is red, the remaining part of the circle is yellow.
Thus, the degree measure of the arc AB (yellow), and therefore the central angle AOB is: 360 0 – 200 0 – 80 0 = 80 0 .
The inscribed angle ACB is half the size of the central angle AOB, that is, equal to 40 degrees.
Answer: 40
What is the inscribed angle subtended by the diameter of the circle? Give your answer in degrees.
It is necessary to know the property of an inscribed angle; understand when and how to use the cosine theorem, learn more about it.
That's all! Good luck to you!
Sincerely, Alexander Krutitskikh
Mathematics teacher at school in the third grade:
- Children, tell me, how much is 6*6?
The children answer in unison:
- Seventy six!
- Well, what are you saying, kids! Six by six will be thirty-six... well, maybe another 37, 38, 39... well, maximum 40... but not seventy-six!
P.S: I would be grateful if you tell me about the site on social networks.
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 the half of the arc on which it subtends.
This should be understood this way: an inscribed angle contains as many angular degrees, minutes and seconds as there are arc degrees, minutes and seconds contained in the half of the arc on which it rests.
When proving this theorem, three cases must be considered.
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 the circle O lies on side BC. It is required to prove that it is measured by half an arc AC.
Connect point A to the center of the circle. We obtain an 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, then ∠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. We need to prove that ∠ABD is measured by half the arc AD.
To prove this, let us draw the diameter BC. Angle ABD is split into two angles: ∠1 and ∠2.
∠1 is measured by half an arc AC, and ∠2 is measured by half an arc CD, therefore, the entire ∠ABD is measured by 1 / 2 \(\breve(AC)\) + 1 / 2 \(\breve(CD)\), i.e. . half 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. We need to prove that ∠MAD is measured by half the arc MD.
To prove this, let's 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 subtending 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 subtended by a diameter is a right angle, since it subtends half a circle. Half a circle contains 180 arc degrees, which means that the angle based on the diameter contains 90 arc degrees (Fig. 334, b).
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 can 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.” It is not always the case that a school textbook is 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.
Central angle is an angle whose vertex is at the center of the circle.
Inscribed angle- an angle whose vertex lies on a circle and whose sides intersect it.
The figure shows central and inscribed angles, as well as their most important properties.
So, the magnitude of the central angle is equal to the angular magnitude of the arc on which it rests. This means that a central angle of 90 degrees will rest on an arc equal to 90°, that is, a circle. The central angle, equal to 60°, rests on an arc of 60 degrees, that is, on the sixth part of the circle.
The magnitude of the inscribed angle is two times less than the central angle based on the same arc.
Also, to solve problems we will need the concept of “chord”.
Equal central angles subtend equal chords.
1. What is the inscribed angle subtended by the diameter of the circle? Give your answer in degrees.
An inscribed angle subtended by a diameter is a right angle.
2. The central angle is 36° greater than the acute inscribed angle subtended by the same circular arc. Find the inscribed angle. Give your answer in degrees.
Let the central angle be equal to x, and the inscribed angle subtended by the same arc be equal to y.
We know that x = 2y.
Hence 2y = 36 + y,
y = 36.
3. The radius of the circle is equal to 1. Find the value of the obtuse inscribed angle subtended by the chord, equal to . Give your answer in degrees.
Let the chord AB be equal to . The 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 already encountered such triangles. Obviously, triangle AOB is rectangular and isosceles, that is, angle AOB is 90°.
Then the arc ACB is equal to 90°, and the arc AKB is equal to 360° - 90° = 270°.
The inscribed angle α rests on the arc AKB and is equal to half the angular value of this arc, that is, 135°.
Answer: 135.
4. The chord AB divides the circle into two parts, the degree values of which are in the ratio 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 conditions. 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 is happening on the chord AB. It’s as if the chord AB is a screen in a movie theater :-)
Obviously, you need to find the angle ACB.
The sum of the two arcs into which the chord AB divides the circle is equal to 360°, that is
5x + 7x = 360°
Hence x = 30°, and then the inscribed angle ACB rests on an arc equal to 210°.
The magnitude of the inscribed angle is equal to half the angular magnitude of the arc on which it rests, which means that angle ACB is equal to 105°.
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. Sine of its half equal to the ratio 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 respectively be 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 is zero, then the arcsine of it is also zero; - of 1/2 is equal to 30° or 1/6 Pi, if measured; - arcsine of -1/2 is -30° or -1/ 6 from the number Pi in; - the arcsine of 1 is equal to 90° or 1/2 of the number Pi in radians; - the arcsine of -1 is equal to -90° or -1/2 of the number 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 operating system used).
Enter the value of the argument from which the arctangent should be calculated. This can be done by clicking the calculator interface buttons with the mouse, or by pressing the keys on , or by copying the value (CTRL + C) and then pasting it (CTRL + V) into the calculator input field.
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. Degree measure 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 altitude drawn from the vertex of the central angle to the side of the triangle formed by the 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
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=√(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
By substituting a numerical value for the value of 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:
Area of the sector corresponding given angle can be calculated using the following formula: