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.
Let's 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.” Usually, this section planimetry has been studied since high school. It is not surprising that many students are faced with the need to review 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.” 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 to the maximum extent possible. 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 “Catalogue” 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 smaller 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°.
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:
- AB is the chord of the circle;
- Point O is the center of the circle, so angle AOB is the central angle;
- 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 of the 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?
- The midpoint of the hypotenuse is equidistant from all the vertices of the triangle. This is a direct consequence of the theorem;
- 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 why angle is equal A. 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.
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