The simplest trigonometric equations. Funny incident from life On the unit circle there are two diametrically opposed

+ – 0;2 P; 4 P. - 2 P; -4 P. P -11 P 6 P -7 P 4 P -5 P 3 2 P -4 P 3 3 P -4 P P -7 P P -5 P P -3 P P -2 P P - P P - P P - P P 2 5 P 2 P 2 9 P 2 5 P 2 P 2 11 P 2 7 P 2 3 P 2 11 P 2 7 P 2 3 P 2 5 P;3 P; P. -5 P;-3 P;- P. 360° 30° 60° 45° 90° 120° 135° 150° 180° 210° 225° 240° 270° 300° 315° 330° X y 0

0 y X 5 P,14 -P-P ± P 2P 2 ± P P k, k Z (-1) k P 4P 4 + P g, g Z P 3P 3 ± + 2 P n, n Z P 6P 6 + P 3P 3 m, m Z Find the points corresponding to the following numbers

0 y X - P +2 P k, k Z P 3P P n, n Z P m, m Z P (+ m), m Z 2P 32P P n, n Z P 2P 2 P P n, n Z 1 3 P (+2 l), l Z Find the points corresponding to the following numbers

1. Which quarter of the number circle does point A belong to? First. B. Second. V. Third. G. Fourth. 2. Which quarter of the number circle does point A belong to? First. B. Second. V. Third. G. Fourth. 3. Determine the signs of the numbers a and b if: A. a>0, b>0. B. a 0. B. a>0, b0, b 0"> 0, b>0. B. a 0. B. a>0, b0, b"> 0" title="1.Which quarter of the number circle does point A. First. B. Second. C. Third. D. Fourth. 2. Which quarter of the number circle does point A. First. B. Second. C. Third. D. Fourth belong to? 3. Determine the signs of the numbers a and b if: A. a>0"> title="1. Which quarter of the number circle does point A belong to? First. B. Second. V. Third. G. Fourth. 2. Which quarter of the number circle does point A belong to? First. B. Second. V. Third. G. Fourth. 3. Determine the signs of the numbers a and b if: A. a>0"> !}

Question: On a circle, diametrically opposite points A and B and a different point C are chosen. The tangent drawn to the circle at point A and the line BC intersect at point D. Prove that the tangent drawn to the circle at point C bisects the segment A.D. The incircle of triangle ABC touches sides AB and BC at points M and N respectively. A line passes through the midpoint of AC parallel to the line. MN intersects lines BA and BC at points D and E, respectively. Prove that AD=CE.

On the circle, diametrically opposite points A and B and a different point C are chosen. The tangent drawn to the circle at point A and the straight line BC intersect at point D. Prove that the tangent drawn to the circle at point C bisects the segment AD. The incircle of triangle ABC touches sides AB and BC at points M and N respectively. A line passes through the midpoint of AC parallel to the line. MN intersects lines BA and BC at points D and E, respectively. Prove that AD=CE.

Answers:

Basic principles of spherical geometry.

Any plane intersecting a sphere produces a circle in cross-section. If the plane passes through the center of the sphere, then the cross section results in a so-called great circle. Through any two points on a sphere, except those that are diametrically opposite, a single large circle can be drawn. (On the globe, an example of a great circle is the equator and all the meridians.) An infinite number of great circles pass through diametrically opposite points. Lesser arc AmB(Fig. 1) of the great circle is the shortest of all lines on the sphere connecting given points. This line is called geodetic. Geodesic lines play the same role on a sphere as straight lines do in planimetry. Many provisions of geometry on the plane are also valid on the sphere, but, unlike the plane, two spherical lines intersect at two diametrically opposite points. Thus, the concept of parallelism simply does not exist in spherical geometry. Another difference is that the spherical line is closed, i.e. moving along it in the same direction, we will return to the starting point; the point does not split the line into two parts. And another surprising fact from the point of view of planimetry is that a triangle on a sphere can have all three right angles.

Lines, segments, distances and angles on a sphere.

Great circles on a sphere are considered to be straight lines. If two points belong to a great circle, then the length of the smaller of the arcs connecting these points is defined as spherical distance between these points, and the arc itself is like a spherical segment. Diametrically opposite points are connected by an infinite number of spherical segments - large semicircles. The length of a spherical segment is determined through the radian measure of the central angle a and the radius of the sphere R(Fig. 2), according to the arc length formula it is equal to R a. Any point WITH spherical segment AB splits it into two, and the sum of their spherical lengths, as in planimetry, is equal to the length of the entire segment, i.e. R AOC+ R OWL= P AOB. For any point D outside the segment AB there is a “spherical triangle inequality”: the sum of the spherical distances from D before A and from D before IN more AB, i.e. R AOD+ R DOB> R AOB, – complete correspondence between spherical and flat geometries. The triangle inequality is one of the fundamental ones in spherical geometry; it follows from it that, as in planimetry, a spherical segment is shorter than any spherical broken line, and therefore any curve on the sphere connecting its ends.

In the same way, many other concepts of planimetry can be transferred to the sphere, in particular those that can be expressed through distances. For example, spherical circle– a set of points on the sphere equidistant from a given point R. It is easy to show that the circle lies in a plane perpendicular to the diameter of the sphere RR` (Fig. 3), i.e. this is an ordinary flat circle with a center on the diameter RR`. But it has two spherical centers: R And R`. These centers are usually called poles. If we turn to the globe, we can see that we are talking about circles such as parallels, and the spherical centers of all parallels are the North and South Poles. If the diameter r of a spherical circle is equal to p/2, then the spherical circle turns into a spherical straight line. (On the globe there is the equator). In this case, such a circle is called polar each of the points R And P`.

One of the most important concepts in geometry is the equality of figures. Figures are considered equal if one can be displayed on top of the other in such a way (by rotation and translation) that the distances are preserved. This is also true for spherical geometry.

Angles on a sphere are defined as follows. When two spherical lines intersect a And b Four spherical bigons are formed on the sphere, just as two intersecting lines on a plane divide it into four plane angles (Fig. 4). Each of the diagons corresponds to a dihedral angle formed by the diametrical planes containing a And b. And the angle between spherical straight lines is equal to the smaller of the angles of the diagons they form.

We also note that angle P ABC, formed on a sphere by two arcs of a great circle, is measured by angle P A`B.C.` between tangents to the corresponding arcs at a point IN(Fig. 5) or a dihedral angle formed by diametrical planes containing spherical segments AB And Sun.

In the same way as in stereometry, each point on the sphere is associated with a ray drawn from the center of the sphere to this point, and any figure on the sphere is associated with the union of all rays intersecting it. Thus, a spherical straight line corresponds to the diametrical plane containing it, a spherical segment corresponds to a plane angle, a digon corresponds to a dihedral angle, and a spherical circle corresponds to a conical surface whose axis passes through the poles of the circle.

A polyhedral angle with a vertex at the center of the sphere intersects the sphere along a spherical polygon (Fig. 6). This is an area on a sphere bounded by a broken line of spherical segments. The links of the broken line are the sides of a spherical polygon. Their lengths are equal to the values of the corresponding plane angles of the polyhedral angle, and the value of the angle at any vertex A equal to the dihedral angle at the edge OA.

Spherical triangle.

Among all spherical polygons, the spherical triangle is of greatest interest. Three large circles, intersecting in pairs at two points, form eight spherical triangles on the sphere. Knowing the elements (sides and angles) of one of them, it is possible to determine the elements of all the others, so we consider the relationships between the elements of one of them, the one whose all sides are less than half of the great circle. The sides of a triangle are measured by the plane angles of the trihedral angle OABC, the angles of the triangle are dihedral angles of the same trihedral angle (Fig. 7).

Many properties of a spherical triangle (and they are also properties of trihedral angles) almost completely repeat the properties of an ordinary triangle. Among them is the triangle inequality, which, in the language of trihedral angles, states that any plane angle of a trihedral angle is less than the sum of the other two. Or, for example, three signs of equality of triangles. All planimetric consequences of the mentioned theorems, together with their proofs, remain valid on the sphere. Thus, the set of points equidistant from the ends of the segment will also be on the sphere perpendicular to it, a straight line passing through its middle, from which it follows that the bisectors are perpendicular to the sides of a spherical triangle ABC have a common point, or rather, two diametrically opposed common points R And R`, which are the poles of its only circumscribed circle (Fig. 8). In stereometry, this means that a cone can be described around any trihedral angle. It is easy to transfer to the sphere the theorem that the bisectors of a triangle intersect at the center of its incircle.

Theorems on the intersection of heights and medians also remain true, but their usual proofs in planimetry directly or indirectly use parallelism, which does not exist on a sphere, and therefore it is easier to prove them again, in the language of stereometry. Rice. Figure 9 illustrates the proof of the spherical median theorem: planes containing the medians of a spherical triangle ABC, intersect a plane triangle with the same vertices along its usual medians, therefore, they all contain the radius of the sphere passing through the intersection point of the plane medians. The end of the radius will be the common point of the three “spherical” medians.

The properties of spherical triangles differ in many ways from the properties of triangles on a plane. Thus, to the known three cases of equality of rectilinear triangles, a fourth is added: two triangles ABC And А`В`С` are equal if three angles P are equal, respectively A= P A`, R IN= P IN`, R WITH= P WITH`. Thus, there are no similar triangles on the sphere; moreover, in spherical geometry there is no very concept of similarity, because There are no transformations that change all distances by the same (not equal to 1) number of times. These features are associated with a violation of the Euclidean axiom of parallel lines and are also inherent in Lobachevsky’s geometry. Triangles that have equal elements and different orientations are called symmetrical, such as triangles AC`WITH And VSS` (Fig. 10).

The sum of the angles of any spherical triangle is always greater than 180°. Difference P A+P IN+P WITH - p = d (measured in radians) is a positive quantity and is called spherical excess of a given spherical triangle. Area of a spherical triangle: S = R 2 d where R is the radius of the sphere, and d is the spherical excess. This formula was first published by the Dutchman A. Girard in 1629 and named after him.

If we consider a diagon with angle a, then at 226 = 2p/ n (n – integer) the sphere can be cut exactly into P copies of such a diagon, and the area of the sphere is 4 nR 2 = 4p at R= 1, so the area of the diagon is 4p/ n= 2a. This formula is also true for a = 2p t/n and therefore true for all a. If we continue the sides of a spherical triangle ABC and express the area of the sphere through the areas of the resulting bigons with angles A,IN,WITH and its own area, then we can arrive at the above Girard formula.

Coordinates on the sphere.

Each point on the sphere is completely determined by specifying two numbers; these numbers ( coordinates) are determined as follows (Fig. 11). Some large circle is fixed QQ` (equator), one of the two points of intersection of the diameter of the sphere PP`, perpendicular to the equatorial plane, with the surface of a sphere, for example R (pole), and one of the large semicircles PAP` coming out of the pole ( first meridian). Large semicircles coming out of P, called meridians, small circles parallel to the equator, such as LL`, – parallels. As one of the point coordinates M on the sphere the angle q is taken = POM (point height), as the second – angle j = AON between the first meridian and the meridian passing through the point M (longitude points, counted counterclockwise).

In geography (on the globe), it is customary to use the Greenwich meridian as the first meridian, passing through the main hall of the Greenwich Observatory (Greenwich is a London borough), it divides the Earth into the Eastern and Western hemispheres, respectively, and longitude is eastern or western and measured from 0 to 180° in both directions from Greenwich. And instead of the height of a point in geography, it is customary to use latitude at, i.e. corner NOM = 90° – q, measured from the equator. Because Since the equator divides the Earth into the Northern and Southern Hemispheres, the latitude is either northern or southern and varies from 0 to 90°.

Marina Fedosova

Final work in MATHEMATICS
Grade 10
April 28, 2017
Option MA00602
(a basic level of)
Completed by: Full name_______________________________________ class ______
Instructions for performing the work
You are given 90 minutes to complete the final math work. Job
includes 15 tasks and consists of two parts.
The answer in the tasks of the first part (1-10) is an integer,
decimal fraction or sequence of numbers. Write your answer in the field
answer in the text of the work.
In task 11 of the second part you need to write down the answer in a special
the field allocated for this.
In tasks 12-14 of the second part you need to write down the solution and answer
in the field provided for this purpose. The answer to task 15 is
function graph.
Each of tasks 5 and 11 is presented in two versions, of which
You only need to select and execute one.
When performing work, you cannot use textbooks, work
notebooks, reference books, calculator.
If necessary, you can use a draft. Entries in draft will not be reviewed or graded.
You can complete tasks in any order, the main thing is to do it correctly
solve as many tasks as possible. We advise you to save time
skip a task that cannot be completed immediately and move on
to the next. If after completing all the work you still have time,
You will be able to return to missed tasks.
We wish you success!

Part 1
In tasks 1-10, give your answer as a whole number, decimal fraction or
sequences of numbers. Write your answer in the answer field in the text
work.
1

The price for an electric kettle was increased by 10% and amounted to
1980 rubles. How many rubles did the kettle cost before the price increase?

Oleg and Tolya left school at the same time and went home in the same direction.
Expensive. The boys live in the same house. The figure shows a graph
the movements of each: Oleg - with a solid line, Tolya - with a dotted line. By
the vertical axis shows the distance (in meters), the horizontal axis shows the distance
travel time for each in minutes.

Using the graph, choose the correct statements.
1)
2)
3)

Oleg came home before Tolya.
Three minutes after leaving school, Oleg caught up with Tolya.
Throughout the entire journey, the distance between the boys was less
100 meters.
4) In the first six minutes the boys covered the same distance.

Answer: ___________________________

Find the meaning of the expression

π
π
- 2 sin 2.
8
8

Answer: ___________________________
StatGrad 2016−2017 academic year. Publishing online or in print
without the written consent of StatGrad it is prohibited

Mathematics. Grade 10. Option 00602 (basic level)

There are two marked on the unit circle
diametrically opposite points Pα and
Pβ corresponding to rotations through angles α and
β (see figure).
Is it possible to say that:
1) α  β  0
2) cosα  cosβ
3) α  β  2π
4) sin α  sin β  0

In your answer, indicate the numbers of the correct statements without spaces, commas and
other additional characters.
Answer: ___________________________
Select and complete only ONE of tasks 5.1 or 5.2.
5.1

The figure shows a graph
function y  f (x) defined on the interval   3;11 .
Find the smallest value
functions on the segment  1; 5.

Answer: ___________________________
5.2

Solve the equation log 2 4 x5  6.

Answer: ___________________________

StatGrad 2016−2017 academic year. Publishing online or in print
without the written consent of StatGrad it is prohibited

Mathematics. Grade 10. Option 00602 (basic level)

A plane passing through points A, B and C (see.
figure), splits the cube into two polyhedra. One of
it has four sides. How many faces does the second one have?

Answer: ___________________________
7

Choose the numbers of the correct statements.
1)
2)
3)
4)

In space, through a point not lying on a given line, you can
draw a plane that does not intersect a given line, and, moreover, only
one.
An inclined line drawn to a plane forms the same angle with
all straight lines lying in this plane.
A plane can be drawn through any two intersecting lines.
Through a point in space that does not lie on a given line, one can
Draw two straight lines that do not intersect a given line.

In your answer, indicate the numbers of the correct statements without spaces, commas and
other additional characters.
Answer: ___________________________
8

On the poultry farm there are only chickens and ducks, and there are 7 times more chickens than
ducks Find the probability that a randomly selected farm
the bird turns out to be a duck.
Answer: ___________________________

The roof of the canopy is located at an angle of 14
to the horizontal. Distance between two supports
is 400 centimeters. Using the table,
determine how many centimeters one support is
longer than the other.
α
13
14
15
16
17
18
19

Sin α
0,225
0,241
0,258
0,275
0,292
0,309
0,325

Cos α
0,974
0,970
0,965
0,961
0,956
0,951
0,945

Tg α
0,230
0,249
0,267
0,286
0,305
0,324
0,344

Answer: ___________________________
StatGrad 2016−2017 academic year. Publishing online or in print
without the written consent of StatGrad it is prohibited

Mathematics. Grade 10. Option 00602 (basic level)

Find the smallest natural seven-digit number that is divisible by 3,
but not divisible by 6 and each digit of which, starting from the second, is less
previous one.
Answer: ___________________________
Part 2
In task 11, write your answer in the space provided. In tasks
12-14 you need to write down the solution and answer in the specially designated space
for this field. The answer to task 15 is the graph of the function.
Select and complete only ONE of the tasks: 11.1 or 11.2.

2
. Write down three different possible values
2
such angles. Give your answer in radians.

Find the smallest natural number that is greater than log 7 80.

The cosine of the angle is 

StatGrad 2016−2017 academic year. Publishing online or in print
without the written consent of StatGrad it is prohibited

Mathematics. Grade 10. Option 00602 (basic level)

In triangle ABC, sides AB and BC are marked
points M and K, respectively, so that BM: AB  1: 2, and
BK:BC  2:3. How many times the area of triangle ABC?
greater than the area of triangle MVK?

Choose some pair of numbers a and b so that the inequality ax  b  0
satisfied exactly three of the five points marked in the figure.
-1

StatGrad 2016−2017 academic year. Publishing online or in print
without the written consent of StatGrad it is prohibited

Mathematics. Grade 10. Option 00602 (basic level)

The price of the iron was increased twice by the same percentage. On
how many percent did the price of the iron increase each time if it
the initial cost is 2000 rubles, and the final cost is 3380 rubles?

StatGrad 2016−2017 academic year. Publishing online or in print
without the written consent of StatGrad it is prohibited

Mathematics. Grade 10. Option 00602 (basic level)

The function y  f (x) has the following properties:
1) f (x)  3 x  4 at 2  x  1;
2) f (x)  x  2 at 1  x  0;
3) f (x)  2  2 x at 0  x  2;
4) the function y  f (x) is periodic with period 4.
Draw a graph of this function on the segment  6;4.
y

StatGrad 2016−2017 academic year. Publishing online or in print
without the written consent of StatGrad it is prohibited