The video course “Get an A” includes all the topics necessary to successfully pass the Unified State Exam in mathematics with 60-65 points. Completely all tasks 1-13 of the Profile Unified State Exam in mathematics. Also suitable for passing the Basic Unified State Examination in mathematics. If you want to pass the Unified State Exam with 90-100 points, you need to solve part 1 in 30 minutes and without mistakes!

Preparation course for the Unified State Exam for grades 10-11, as well as for teachers. Everything you need to solve Part 1 of the Unified State Exam in mathematics (the first 12 problems) and Problem 13 (trigonometry). And this is more than 70 points on the Unified State Exam, and neither a 100-point student nor a humanities student can do without them.

All the necessary theory. Quick solutions, pitfalls and secrets of the Unified State Exam. All current tasks of part 1 from the FIPI Task Bank have been analyzed. The course fully complies with the requirements of the Unified State Exam 2018.

The course contains 5 large topics, 2.5 hours each. Each topic is given from scratch, simply and clearly.

Hundreds of Unified State Exam tasks. Word problems and probability theory. Simple and easy to remember algorithms for solving problems. Geometry. Theory, reference material, analysis of all types of Unified State Examination tasks. Stereometry. Tricky solutions, useful cheat sheets, development of spatial imagination. Trigonometry from scratch to problem 13. Understanding instead of cramming. Clear explanations of complex concepts. Algebra. Roots, powers and logarithms, function and derivative. A basis for solving complex problems of Part 2 of the Unified State Exam.

Proof:

• Given triangle ABC.
• Through vertex B we draw a straight line DK parallel to the base AC.
• \angle CBK= \angle C as internal crosswise lying with parallel DK and AC, and secant BC.
• \angle DBA = \angle A internal crosswise lying with DK \parallel AC and secant AB. Angle DBK is reversed and equal to
• \angle DBK = \angle DBA + \angle B + \angle CBK
• Since the unfolded angle is equal to 180 ^\circ , and \angle CBK = \angle C and \angle DBA = \angle A , we get 180 ^\circ = \angle A + \angle B + \angle C.

The theorem is proven

Corollaries from the theorem on the sum of angles of a triangle:

1. The sum of the acute angles of a right triangle is equal to 90°.
2. In an isosceles right triangle, each acute angle is equal to 45°.
3. In an equilateral triangle, each angle is equal 60°.
4. In any triangle, either all the angles are acute, or two angles are acute, and the third is obtuse or right.
5. An exterior angle of a triangle is equal to the sum of two interior angles that are not adjacent to it.

Triangle Exterior Angle Theorem

An exterior angle of a triangle is equal to the sum of the two remaining angles of the triangle that are not adjacent to this exterior angle

Proof:

• Given a triangle ABC, where BCD is the exterior angle.
• \angle BAC + \angle ABC +\angle BCA = 180^0
• From the equalities the angle \angle BCD + \angle BCA = 180^0
• We get \angle BCD = \angle BAC+\angle ABC.

Preliminary information

First, let's look directly at the concept of a triangle.

Definition 1

We will call a triangle a geometric figure that is made up of three points connected to each other by segments (Fig. 1).

Definition 2

Within the framework of Definition 1, we will call the points the vertices of the triangle.

Definition 3

Within the framework of Definition 1, the segments will be called sides of the triangle.

Obviously, any triangle will have 3 vertices, as well as three sides.

Theorem on the sum of angles in a triangle

Let us introduce and prove one of the main theorems related to triangles, namely the theorem on the sum of angles in a triangle.

Theorem 1

The sum of the angles in any arbitrary triangle is $180^\circ$.

Proof.

Consider the triangle $EGF$. Let us prove that the sum of the angles in this triangle is equal to $180^\circ$. Let's make an additional construction: draw the straight line $XY||EG$ (Fig. 2)

Since the lines $XY$ and $EG$ are parallel, then $∠E=∠XFE$ lie crosswise at the secant $FE$, and $∠G=∠YFG$ lie crosswise at the secant $FG$

Angle $XFY$ will be reversed and therefore equals $180^\circ$.

$∠XFY=∠XFE+∠F+∠YFG=180^\circ$

Hence

$∠E+∠F+∠G=180^\circ$

The theorem has been proven.

Triangle Exterior Angle Theorem

Another theorem on the sum of angles for a triangle can be considered the theorem on the external angle. First, let's introduce this concept.

Definition 4

We will call an external angle of a triangle an angle that will be adjacent to any angle of the triangle (Fig. 3).

Let us now consider the theorem directly.

Theorem 2

An exterior angle of a triangle is equal to the sum of two angles of the triangle that are not adjacent to it.

Proof.

Consider an arbitrary triangle $EFG$. Let it have an external angle of the triangle $FGQ$ (Fig. 3).

By Theorem 1, we will have that $∠E+∠F+∠G=180^\circ$, therefore,

$∠G=180^\circ-(∠E+∠F)$

Since the angle $FGQ$ is external, it is adjacent to the angle $∠G$, then

$∠FGQ=180^\circ-∠G=180^\circ-180^\circ+(∠E+∠F)=∠E+∠F$

The theorem has been proven.

Sample tasks

Example 1

Find all angles of a triangle if it is equilateral.

Since all the sides of an equilateral triangle are equal, we will have that all the angles in it are also equal to each other. Let us denote their degree measures by $α$.

Then, by Theorem 1 we get

$α+α+α=180^\circ$

Answer: all angles equal $60^\circ$.

Example 2

Find all angles of an isosceles triangle if one of its angles is equal to $100^\circ$.

Let us introduce the following notation for angles in an isosceles triangle:

Since we are not given in the condition exactly what angle $100^\circ$ is equal to, then two cases are possible:

An angle equal to $100^\circ$ is the angle at the base of the triangle.

Using the theorem on angles at the base of an isosceles triangle, we obtain

$∠2=∠3=100^\circ$

But then only their sum will be greater than $180^\circ$, which contradicts the conditions of Theorem 1. This means that this case does not occur.

An angle equal to $100^\circ$ is the angle between equal sides, that is

Following on from yesterday:

Let's play with a mosaic based on a geometry fairy tale:

Once upon a time there were triangles. So similar that they are just copies of each other.
They somehow stood side by side in a straight line. And since they were all the same height -
then their tops were at the same level, under the ruler:



Triangles loved to tumble and stand on their heads. They climbed to the top row and stood on the corner like acrobats.
And we already know - when they stand with their tops exactly in a line,
then their soles also follow a ruler - because if someone is the same height, then they are also the same height upside down!



They were the same in everything - the same height, and the same soles,
and the slides on the sides - one steeper, the other flatter - are the same in length
and they have the same slope. Well, just twins! (only in different clothes, each with their own piece of the puzzle).



- Where do the triangles have identical sides? Where are the corners the same?

The triangles stood on their heads, stood there, and decided to slide off and lie down in the bottom row.
They slid and slid down a hill; but their slides are the same!
So they fit exactly between the lower triangles, without gaps, and no one pushed anyone aside.



We looked around the triangles and noticed an interesting feature.
Wherever their angles come together, all three angles will certainly meet:
the largest is the “head angle”, the most acute angle and the third, medium largest angle.
They even tied colored ribbons so that it would be immediately obvious which was which.

And it turned out that the three angles of the triangle, if you combine them -
make up one large angle, an “open corner” - like the cover of an open book,

______________________O ___________________

it's called a turned angle.

Any triangle is like a passport: three angles together are equal to the unfolded angle.
Someone knocks on your door: - knock-knock, I'm a triangle, let me spend the night!
And you tell him - Show me the sum of the angles in expanded form!
And it’s immediately clear whether this is a real triangle or an impostor.
Failed verification - Turn around one hundred and eighty degrees and go home!



When they say "turn 180°" it means to turn around backwards and
go in the opposite direction.

The same thing in more familiar expressions, without “once upon a time”:



Let us perform a parallel translation of triangle ABC along the OX axis
to vector AB equal to the length of the base AB.
Line DF passing through vertices C and C 1 of triangles
parallel to the OX axis, due to the fact that perpendicular to the OX axis
segments h and h 1 (heights of equal triangles) are equal.
Thus, the base of the triangle A 2 B 2 C 2 is parallel to the base AB
and equal to it in length (since the vertex C 1 is shifted relative to C by the amount AB).
Triangles A 2 B 2 C 2 and ABC are equal on three sides.
Therefore, the angles ∠A 1 ∠B ∠C 2 forming a straight angle are equal to the angles of triangle ABC.
=> The sum of the angles of a triangle is 180°

With movements - “translations”, the so-called proof is shorter and clearer,
even a child can understand the pieces of the mosaic.

But traditional school:



based on the equality of internal cross-lying angles cut off on parallel lines



valuable in that it gives an idea of ​​why this is so,
Why the sum of the angles of a triangle is equal to the reverse angle?

Because otherwise parallel lines would not have the properties familiar to our world.

The theorems work both ways. From the axiom of parallel lines it follows
equality of crosswise lying and vertical angles, and from them - the sum of the angles of a triangle.

But the opposite is also true: as long as the angles of a triangle are 180°, there are parallel lines
(such that through a point not lying on a line one can draw a unique line || of the given one).
If one day a triangle appears in the world whose sum of angles is not equal to the unfolded angle -
then the parallel ones will cease to be parallel, the whole world will be bent and skewed.

If stripes with triangle patterns are placed one above the other -
you can cover the entire field with a repeating pattern, like a floor with tiles:




you can trace different shapes on such a grid - hexagons, rhombuses,
star polygons and get a variety of parquets






Tiling a plane with parquet is not only an entertaining game, but also a relevant mathematical problem:



________________________________________ _______________________-------__________ ________________________________________ ______________
/\__||_/\__||_/\__||_/\__||_/\__|)0(|_/\__||_/\__||_/\__||_/\__||_/\=/\__||_/ \__||_/\__||_/\__||_/\__|)0(|_/\__||_/\__||_/\__||_/\__||_/\

Since every quadrilateral is a rectangle, square, rhombus, etc.,
can be composed of two triangles,
respectively, the sum of the angles of a quadrilateral: 180° + 180° = 360°


Identical isosceles triangles are folded into squares in different ways.
A small square of 2 parts. Average of 4. And the largest of the 8.
How many figures are there in the drawing, consisting of 6 triangles?