An angle formed by two chords drawn from the same point is called an inscribed angle.
THEOREM An inscribed angle is measured by half the arc it intercepts.
Consequences:
all inscribed angles based on the same arc are equal;
An inscribed angle based on a diameter is a right angle.
THEOREM An angle whose vertex lies inside a circle is measured by half the sum of two arcs enclosed between its sides
THEOREM An angle whose vertex lies outside the circle and whose sides intersect the circle is measured by the half-difference of the two arcs enclosed between its sides.
THEOREM An angle formed by a tangent and a chord is measured by half the arc contained within the angle.
Tasks with a solution
1. Find the angle ABC. Give your answer in degrees.
Decision.
Construct a square with side AC.
Then it can be seen that the angle ABC is based on circles, that is, on an arc of 90º. An inscribed angle is half the arc it intercepts, so 
2. The chord AB divides the circle into two parts, the degree values of which are related as 6:12. At what angle is this chord visible from point C, which belongs to the smaller arc of the circle? Give your answer in degrees.
Decision.
From a point C chord AB seen at an angle ACB. Let the largest part of the circle be 12x, then the smaller one is 6x. The whole circle is 360º.
We get the equation 12x + 6x \u003d 360º. From where x \u003d 20º.
Injection DIA rests on a large arc of a circle, which is equal to 12 20º=240º.
An inscribed angle is equal to half the arc on which it rests, which means that the angle resting on a large arc ACB equals
Answer 120º
3. Chord AB subtends the arc of a circle at 84º. Find an angle ABC between this chord and the tangent to the circle through point B. Give your answer in degrees.
Decision.
Injection ABC is the angle between the tangent and the chord. It is measured by half the arc enclosed inside the corner. The arc inside the angle is 84º. So 
4. A tangent is drawn to a circle of radius 36 from a point remote from the center by a distance equal to 85. Find the length of the tangent.
Let OA=36, OS=85. The radius drawn to the point of contact is perpendicular to the tangent. From the right triangle AOC, by the Pythagorean theorem, we obtain
5. To a circle from a point With tangent drawn outside it AC and secant CD, intersecting circle at a point AT. The sum of the lengths of the tangent and secant is 30 cm, and the inner segment of the secant is 2 cm shorter than the tangent. Find the lengths of the tangent and secant.
Let be AC=x and CD=y. Then x+y=30, and DB=AC-2=x-2 , BC=AC-DB=y-DB=y-(x-2)=y-x+2. According to the theorem, if a tangent and a secant are drawn to it from a point outside the circle, then the square of the tangent is equal to the product of the secant by its outer part, that is 
. Then 
We get the system 
. X=80 is not suitable because at>0 Therefore, we get
Tangent AC=12, secant CD=18.
Answer 12 and 18
6. Find the area S of the shaded sector. Give your answer S/π.
Let's build a square on this drawing
Then it becomes obvious that the sector is one quarter of the circle.
The radius is half the diagonal of a square whose side is 4.
Then we calculate the area of the sector by the formula
Then the desired value is equal to
| What is the inscribed angle based on the diameter of the circle? Give your answer in degrees. | Find the chord on which the angle 90º rests, inscribed in a circle of radius 1. |
What is an acute inscribed angle that intercepts a chord equal to the radius of the circle? Give your answer in degrees.  | Find the chord on which the angle of 30º rests, inscribed in a circle of radius 3.  |
What is an obtuse inscribed angle subtended by a chord equal to the radius of the circle? Give your answer in degrees.  | The radius of the circle is 1. Find the value of the acute inscribed angle based on the chord equal to . Give your answer in degrees. |
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.  | Find the chord on which the angle 120º rests, inscribed in a circle of radius .  |
The central angle is 34º greater than the acute inscribed angle based on the same circular arc. Find the inscribed angle. Give your answer in degrees.  |  |
Find the angle ABC. Give your answer in degrees.  | Find the degree value of the arc AC of the circle on which the angle ABC rests. Give your answer in degrees.  |
| Find the degree value of the arc BC of the circle on which the angle BAC rests. Give your answer in degrees. | Angle ACO is 25º, where O is the center of the circle. Its side CA touches the circle. Find the magnitude of the smaller arc AB of the circle contained within this angle. Give your answer in degrees.  |
Find the angle ACO if its side CA is tangent to the circle, O is the center of the circle, and the major arc AD of the circle contained within this angle is 110º. Give your answer in degrees.  | Find the angle ACB if the inscribed angles ADB and DAE are based on arcs of a circle whose degree values are 116º and 36º respectively. Give your answer in degrees.  |
Angle ACB is 50º. The degree value of the arc AB of a circle that does not contain points D and E is equal to 130º. Find the angle DAE. Give your answer in degrees.  | Chord AB subtends an arc of a circle at 86º. Find the angle ABC between this chord and the tangent to the circle through point B. Give your answer in degrees.  |
The angle between chord AB and tangent BC to the circle is 28º. Find the magnitude of the smaller arc subtracted by chord AB. Give your answer in degrees.  | Tangents AC and BC are drawn through the ends A, B of a circular arc of 72º. Find angle ACB. Give your answer in degrees.  |
The tangents CA and CB to the circle form an angle ACB equal to 112º. Find the value of the smaller arc AB subtracted by the points of contact. Give your answer in degrees.  | Find the angle ACO if its side CA is tangent to the circle, O is the center of the circle, and the lesser arc of the circle AB contained within this angle is equal to 62º. Give your answer in degrees.  |
Lesson Objectives: formation of knowledge on the topic, organization of work on the assimilation of concepts, scientific facts.
Educational tasks:
- introduce the concept of an inscribed angle;
- teach to recognize inscribed angles in drawings;
- anticipate an additional construction containing an inscribed angle leading to a solution of the problem;
- consider the inscribed angle theorem and its consequences;
- show the application of the theorem in solving problems;
- learn about optical illusions
Educational tasks: activation of independent cognitive activity of students. formation of teamwork skills, development of a sense of responsibility for one's knowledge, culture of communication, familiarization with the knowledge of optical illusion and its application in practice, education of aesthetic culture.
Developmental tasks: to continue the development of the ability to analyze, compare, compare, highlight the main thing, establish cause-and-effect relationships; improve graphic culture.
Technology: problematic study using information technology.
Type of lesson: a lesson in the formation of new knowledge.
Lesson form: lesson - problem statement.
Lesson equipment: presentation: presentation, introspection sheets.
Lesson stages
- Motivation for learning activities -1 minute.
- State the problem and create a plan to solve it - 2 minutes.
- Updating knowledge - 4 minutes.
- Discovery of a new concept - 10 minutes.
- Research work to identify the properties of a new concept - 4 minutes.
- Application of new knowledge - 11 minutes.
- The game "Believe - do not believe" in order to consolidate the new theoretical material - 2 minutes.
- Individual work with the test - 5 minutes.
- Applying new knowledge in unfamiliar situations - 4 minutes.
- Reflection - 3 minutes.
During the classes
1. Motivation for learning activities
Hello guys. Sit down. I hope that the knowledge that you get in the lesson will be useful to you in life.
2. State the problem and create a plan to solve it
Given a flower bed of a round shape, on one of the chords of which roses are planted. In what different places in the flower bed should three rose bushes be planted in such a way that from these points all the roses are visible from the same angle? (Slide 2). Presentation
What solutions do you have for this problem?
A problem situation arises. Students lack knowledge.
To answer this question, you need to use the properties of the inscribed angle. Then let's make a lesson plan together. What are the goals of the lesson and how will we achieve them? During the discussion, the lesson plan appears on the screen. (C lay 3)
3. Updating knowledge
Teacher: Define an angle. What is called the central angle? (C lay 4)
Tasks (Slide 5
4. Discovery of a new concept
Now you see six drawings. What groups would you divide them into and why? (Slide 6)
Sharp, straight, blunt.
Corners 1, 3, 5 and 2, 4, 6 by the location of the corner vertex? What are the angles 1, 3, 5 called?
And angles 2, 4, 6 are called inscribed. That's what we're going to talk about today.
How are angles ABC and KRO similar and how are they different? (Slide 7)
After answering this question, the students try to define the inscribed angle, after which the teacher displays the wording, emphasizing the important points: (C lay 8)
- the vertex lies on the circle,
- sides intersect the circle.
Find pictures that show inscribed angles.
Exercise. Express the value of the inscribed angle, knowing how the value of the central angle is expressed through the arc on which it rests. Working with slide 10
What additional building needs to be done to complete the specified task? If students do not immediately guess, clarify: which central angle should be associated with this inscribed angle?
Further, students see that the resulting central angle is the external angle of an isosceles triangle and come to the conclusion that one of the angles (in particular, inscribed), equal to their half-sum, is equal to half the central one, i.e. half of the arc on which it rests.
An exact formulation of the theorem is given and projected onto a screen. (C lay 11).
Pupils transfer the drawing to the notebook ( slide 12), then write down the condition in the notebook. One of the students comments on the notes. The next student writes down and comments on the proof of the theorem. The consistency and completeness of the design is checked using slide 12). Thus, the proof of the theorem is formalized for the case when the side of the inscribed angle passes through the center of the circle.
The case when the center of the circle lies inside the corner is considered verbally using slide 13.
The next case, when the center of the circle lies outside the corner, the teacher offers to justify it yourself during home preparation. (C lay 14). In the classroom, according to the drawing slide 15 find out that a given inscribed angle can be considered as the difference of two angles, each of which has one side that is any side of the given angle, and the other side is common and passes through the center of the circle.
5. Research work to identify the properties of a new concept
Working with slide 15.
Exercise. How to quickly construct several angles equal to a given angle using a compass and straightedge? They notice that their ways are not rational. A problematic situation arises: old knowledge does not provide a rational solution to the problem.
Think about how, using new material, you can solve this problem. It is possible to draw a circle passing through the vertex of the angle without specifying the center and construct various inscribed angles based on the same arc. The problem situation is resolved. After that, Corollary 1 is formulated: “The inscribed angles based on the same arc are equal.”
The work leading to the formulation of Corollary 2 is carried out similarly. (C lay 16)
How to quickly draw a right angle using a compass and straightedge? It is clarified that “quickly” should be understood as “the minimum number of steps”. We come to the irrationality of this construction. If the students did not guess how to complete the construction, the teacher asks the question: on which arc should the right inscribed angle rest? After that, the students outline the step-by-step construction process:
- Draw a circle of arbitrary radius.
- Draw diameter.
- Select any point on the circle, except for the ends of the diameter.
- Draw rays from the selected point through the ends of the diameter.
After that, the teacher says that in this construction, Corollary 2 from the inscribed angle theorem was used. Try to formulate it.
The revised wording is projected onto the screen. ( Slides 17-19)
6. Applying new knowledge
Solving problems to consolidate new material. Working with slides 20-26.
7. A game of repetition in order to consolidate the theoretical material. (C lay 27)
The game "Believe - do not believe"
- Do you believe that if the value of the central angle is 90˚, then the inscribed angle based on this arc is 45˚?
- Do you believe that the segments of the tangents to the circle are equal and make equal angles with the line passing through the center of the circle? Do you believe that the angle passing through the center of the circle is called its central angle?
- Do you believe that an inscribed angle is measured by half of the arc it spans?
- Do you believe that the magnitude of the central angle is twice the magnitude of the arc on which it rests?
- Do you believe that an inscribed angle based on a semicircle is 180˚?
- Do you believe that an angle whose sides intersect a circle called an inscribed angle?
- Do you believe that inscribed angles based on the same arc are equal?
- Do you believe that with further study of the material, not only angles, but also triangles and quadrangles will be associated with a circle?
8. Individual work with the test. (C lays 28-30)
Answer sheets are handed over to the teacher. The teacher then comments on the solutions.
Option 1.
1. The angle DAB is 38° less than the angle AOB. Find the sum of angles AOB and DAB
a) 96°; b) 114°; c) 104°; d) 76°;
2. MP - diameter, O - center of the circle. OM=OK=MK. Find the RKO angle.
a) 60°; b) 40°; c) 30°; d) 45°;
3. Angle ABC is inscribed, angle AOC is central. Find the angle ABC if the angle AOC=126°
a) 112°; b) 123°; c) 117°; d) 113°;
Option 2.
1. The MSC angle is 34 ° less than the IOC angle. Find the sum of the angles MSC and IOC.
a) 112°; b) 102°; c) 96°; d) 68°;
2. AC is the diameter of the circle, O is its center. AB=OB=OA. Find the angle OBC.
a) 50°; b) 60°; c) 30°; d) 45°;
3. O - the center of the circle, the angle L = 136 °. Find angle B.
a) 292°; b) 224°; c) 112°; d) 146°;
Answers to tasks are checked after filling the test.
| Tasks | 1 | 2 | 3 |
| 1 option | B | AT | AT |
| Option 2 | B | AT | AT |
9. Applying new knowledge in unfamiliar situations
a) Working with slides 31-33.
Teacher: “At home, you solved the problem of calculating the angles of a five-pointed star inscribed in a circle. How did you solve it?"
How to solve this problem using the inscribed angle theorem.
Method II: When the vertices of a pentagonal star divide the circle into equal arcs, the problem is solved very simply: 360°: 5:2 *5=180°.
b) Analysis of mathematical sophism on the application of the theorem on the value of the inscribed angle.
A chord that does not pass through the center is equal to the diameter. (C lay 34-36) Find an error in reasoning.
Decision. Let the diameter AB be drawn in a circle. Through point B we draw some chord BC that does not pass through the center, then through the middle of this chord D and point A we draw a new chord AE. Finally, points E and C are connected by a straight line segment. Consider ▲ABD and ▲EDC. In these triangles: BD = DC (by construction), Ð A = Ð C (as inscribed, based on the same arc). In addition, Ð BDA= Ð EDC (as vertical). If the side and two angles of one triangle are respectively equal to the side and two angles of another triangle, then such triangles are congruent. Means,
▲ BDA = ▲ EDC, and in equal triangles opposite equal angles lie equal sides.
Therefore, AB=EC.
Find an error in reasoning.
c) Test for optical illusion according to drawings with an alternative answer. ( Slides 37-39)
Show what illusory deformation sharp central angles and inscribed angles cause.
Test1. Here the illusory deformation is caused by sharp central angles. Although the angles AOB, BOC, COD are equal, but due to the many sharp angles at which the two angles are broken, they pretend to be larger than the average angle.
Test 2-3. Circles are dominant here. The angles inscribed in a circle form a square in the first case, and a regular triangle in the second. These figures, due to the many circles, give themselves out as figures close to a square and a triangle. The sides appear to be concave inwards.
So, we can apply the illusion in practice, in everyday life. For example, with its help, you can hide the flaws in the shape of the face, figure.
10. Reflection
Let's go back to the lesson plan and see if we answered all the questions?
We have not answered one question. So how should three roses be planted? (Slide 40-41)
Having mastered the theorem on the value of an inscribed angle in a circle, we conclude, because from all points of the circle, except for the ends of the chord, this chord is visible at the same angle, we can plant rose bushes at any point on the circle of the flower bed, except for points M and N. This is one of the practical applications of the theorem on the value of the inscribed angle in a circle.
At the end of the lesson, students can be given a questionnaire to fill out, which allows them to carry out self-analysis, give a qualitative and quantitative assessment of the lesson, while, in addition, a task can be formulated to justify their answer:
1. At the lesson I worked ...;
2. With my work in the lesson, I ...;
3. The lesson seemed to me ...;
4. For the lesson I ...;
5. The lesson material to me was…;
6. Homework seems to me ...
Homework. (C lay 42)
- P. 71, learn the definition of an inscribed angle;
- learn the inscribed angle theorem (by writing down the proof of 3 cases) and two corollaries from it;
- № 654 № 656 № 657.
Bibliography:
- Geometry: Proc. For 7–9 cells. general images. institutions / L.S. Atanasyan, V.F. Butuzov, S.B. Kadomtsev and others - 12th ed., - M .: Education, 2002
- Ziv B.G., Meyler V.M., Didactic materials on geometry for grade 8. – 6th ed. - M .: Education, 2002
- Smirnova I.M., Smirnov V.A. Oral exercises in geometry for grades 7-11. The book for the teacher. M.; Enlightenment, 2003
- Rabinovich E.M. Tasks and exercises on ready-made drawings. Geometry grades 7–9. “Ileksa”, “Gymnasium”, Moscow-Kharkov, 2003
CORs and Internet sites:
- Workshop. Multimedia presentations for math lessons. http://www.intergu.ru/infoteka/
- Internet State of Teachers in Infothek-Mathematics. http://www.intergu.ru/infoteka/
- CERs from the Creative Teachers Network portal.
Inscribed angles Inscribed angle theorem 1 case Ray BO coincides with the side of angle ABC Inscribed angle theorem 1 case Ray BO coincides with angle side ABC AOB is isosceles, since OB \u003d OA \u003d R, which means B \u003d A. 2. COA is an external angle, therefore, COA \u003d OVA + OAB COA \u003d 2 OVA, which means OVA \u003d ½ SOA CBA \u003d ½ AC.
°
Repetition game "Believe it or not" Do you believe that if the value of the central angle is 90˚, then the inscribed angle based on this arc is 45˚? Do you believe that the segments of the tangents to the circle are equal and make equal angles with the line passing through the center of the circle? Do you believe that the angle passing through the center of a circle is called its central angle? Do you believe that an inscribed angle is measured by half of the arc it spans? Do you believe that the magnitude of the central angle is twice the magnitude of the arc on which it rests? Do you believe that an inscribed angle based on a semicircle is 180˚? Do you believe that an angle whose sides intersect a circle is called an inscribed angle? Do you believe that inscribed angles based on the same arc are equal? Do you believe that with further study of the material, not only angles, but also triangles and quadrangles will be associated with a circle? No, the segments of tangents to the circle (drawn from one point) are equal and make equal angles with the line passing through (this point and) the center of the circle. YES, if the value of the central angle is 90˚, then the inscribed angle based on this arc is 45˚. No, the angle passing (coming out) through the center of the circle is called its central angle. Yes, an inscribed angle is measured by half of the arc it spans. No, the value of the central angle is twice as large (equal) as the value of the arc on which it rests. No, the inscribed angle based on the semicircle is 180˚ (right). No, an angle whose sides intersect the circle (and the vertex lies on the circle) is called an inscribed angle. Yes, inscribed angles subtending the same arc are equal. Yes, with further study of the material, not only angles will be associated with a circle, but also triangles and quadrangles.
Inscribed angles Work on the test with programmed solution control. Variant Angle DAB is 38° less than angle AOB. Find the sum of the angles AOB and DAB a) 96 °; b) 114 °; c) 104 °; d) 76°; 2. MP - diameter, O - center of the circle. OM=OK=MK. Find the RKO angle. a) 60°; b) 40°; c) 30°; d) 45°; 3. Angle ABC is inscribed, angle AOC is central. Find the angle ABC if the angle AOC \u003d 126 ° a) 112 °; b) 123°; c) 117°; d) 113°; Variant The MSC angle is 34° less than the IOC angle. Find the sum of the angles MSC and IOC. a) 112°; b) 102°; c) 96°; d) 68°; 2. AC is the diameter of the circle, O is its center. AB=OB=OA. Find the angle OBC. a) 50°; b) 60°; c) 30°; d) 45°; 3. O - the center of the circle, the angle L = 136 °. Find the angle B. a) 292 °; b) 224°; c) 112°; d) 146°;
A chord that does not pass through the center is equal to the diameter. Let the diameter AB be drawn in a circle. Through point B we draw some chord BC not passing through the center, then through the middle of this chord D and point A we draw a new chord AE. Finally, points E and C are connected by a straight line segment. Consider ABD and EDC. In these triangles: BD = DC (by construction), A = C (as inscribed, based on the same arc). In addition, BDA = EDC (as vertical). If the side and two angles of one triangle are respectively equal to the side and two angles of another triangle, then such triangles are congruent. This means that BDA \u003d EDC, and equal sides lie in equal triangles opposite equal angles. Therefore, AB=EC.
Let's find the error According to the triangle equality theorem: If the side and two angles adjacent to it of one triangle are respectively equal to the side and two angles adjacent to it of another triangle, then such triangles are equal. And in our case, angle A is not adjacent to side BD.
Inscribed Angles Optical illusion test based on drawings with an alternative answer. We quite often observe an optical illusion and even use it in our practice, but we know very little about its essence. The illusion of vision is used by architects when constructing buildings, fashion designers when creating models, and artists when creating scenery. We know that a light-colored body appears larger than a dark-colored body of the same size. There are reasons that cause optical illusions. Inscribed angles Test 2 Test 3 Test 2 Test 3 Inscribed in a circle: 1. square 2. figure close to a square Test 2, 3: Circles are dominant here. Angles inscribed in a circle form a square in the first case, and a regular triangle in the second. These figures, due to the many circles, give themselves out as figures close to a square and a triangle. The sides appear to be concave inwards. So, we can apply the illusion in practice, in everyday life. For example, with its help, you can hide the flaws in the shape of the face, figure. Inscribed in a circle: 1. triangle 2. figure close to a triangle
Inscribed angles from all points of the circle, except for the ends of the chord, this chord is visible at the same angle, we can plant rose bushes at any point on the circle of the flower bed, except for points M and N. This is one of the practical applications of the theorem on the value of the inscribed angle in a circle.
Inscribed Angles Homework. p. 71, learn the definition of an inscribed angle; learn the inscribed angle theorem (by writing down the proof of 3 cases) and two corollaries from it;
Angle calculation II
- Angle A of quadrilateral ABCD inscribed in a circle is equal to 126 o . Find angle C of this quadrilateral. Give your answer in degrees.
- The sides of the quadrilateral ABCD AB, BC, CD and AD subtend the arcs of the circumscribed circle, the degree values of which are respectively 63 o , 62 o , 90 o and 145 o . Find angle B of this quadrilateral. Give your answer in degrees.
- Points A, B, C and D, located on a circle, divide this circle into four arcs AB, BC, CD and AD, the degree values of which are related respectively as 1: 4: 12: 19. Find angle A of quadrilateral ABCD. Give your answer in degrees.
- Points A, B, C and D, located on a circle, divide this circle into four arcs AB, BC, CD and AD, the degree values of which are related respectively as 1: 5: 10: 20. Find angle A of quadrilateral ABCD. Give your answer in degrees.
- Quadrilateral ABCD is inscribed in a circle. Angle ABC is 58o, angle CAD is 43o. Find angle ABD. Give your answer in degrees.
- The two angles of a quadrangle inscribed in a circle are 25 o and 51 o . Find the largest of the remaining corners. Give your answer in degrees.
- Angles A, B and C of the quadrilateral ABCD are related as 1: 13: 17. Find the angle D if a circle can be circumscribed around this quadrilateral. Give your answer in degrees.
- The central angle is 45 o greater than the acute inscribed angle based on the same circular arc. Find the inscribed angle. Give your answer in degrees.
- The central angle is 47 o greater than the acute inscribed angle based on the same circular arc. Find the inscribed angle. Give your answer in degrees.
- Find the inscribed angle based on the arc that makes up the circle. Give your answer in degrees.
- Find the inscribed angle based on the arc that is 20% of the circle. Give your answer in degrees.
- Find an inscribed angle based on an arc that is 10% of the circle. Give your answer in degrees.
- The arc of a circle AC, not containing the point B, is 180 o . And the arc of the circle BC, not containing the point A, is 45 o . Find the inscribed angle ACB. Give your answer in degrees.
- Points A, B and C, located on the circle, divide it into three arcs, the degree values of which are related as 1: 4: 13. Find the largest angle of the triangle ABC. Give your answer in degrees.
- AC and BD are the diameters of the circle with center O. The angle DIA is 35 o . Find the angle AOD. Give your answer in degrees.
- AC and BD are the diameters of the circle with center O. The angle DIA is 39 o . Find the angle AOD. Give your answer in degrees.
- The chord AB subtracts the arc of a circle to 6 o. Find the acute angle ABC between this chord and the tangent to the circle through point B. Give your answer in degrees.
- The chord AB subtracts the arc of a circle to 114 o. Find the acute angle ABC between this chord and the tangent to the circle through point B. Give your answer in degrees.
- A circle is inscribed in angle C with a value of 107 o, which touches the sides of the angle at points A and B. Find the angle AOB, where point O is the center of the circle. Give your answer in degrees.
- The tangents at point A and B to the circle with center O intersect at an angle of 2 o . Find the angle ABO. Give your answer in degrees.
- Find the angle CDB if the inscribed angles ADB and ADC are based on circular arcs whose degree values are 67 o and 25 o respectively. Give your answer in degrees.
- The angle between the side of a regular -gon inscribed in a circle and the radius of this circle drawn into one of the vertices of the side is 75 o . Find .
- The angle between the side of a regular -gon inscribed in a circle and the radius of this circle drawn into one of the vertices of the side is 54 o . Find .
- The angle between the side of a regular -gon inscribed in a circle and the radius of this circle drawn into one of the vertices of the side is 30 o . Find .
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°.
