At a certain angle a. From a certain angle. Cleat

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Even the most hardened skeptics believe what their senses tell them, but the senses are easily deceived.

An optical illusion is an impression of a visible object or phenomenon that does not correspond to reality, i.e. optical illusion. Translated from Latin, the word “illusion” means “error, delusion.” This suggests that illusions have long been interpreted as some kind of malfunction in the visual system. Many researchers have been studying the causes of their occurrence.

Some visual illusions have long been scientific explanation, others still remain a mystery.

website continues to collect the coolest optical illusions. Be careful! Some illusions can cause tearing, headaches and disorientation in space.

Endless chocolate

If you cut a chocolate bar 5 by 5 and rearrange all the pieces in the order shown, then out of nowhere an extra piece of chocolate will appear. You can do the same with an ordinary chocolate bar and make sure that this is not computer graphics, but a real-life riddle.

Illusion of bars

Take a look at these bars. Depending on which end you are looking at, the two pieces of wood will either be next to each other, or one of them will be lying on top of the other.

Cube and two identical cups

Optical illusion created by Chris Westall. There is a cup on the table, next to which there is a cube with a small cup. However, upon closer examination, we can see that in fact the cube is drawn, and the cups are exactly the same size. A similar effect is noticeable only at a certain angle.

Illusion "Cafe Wall"

Take a close look at the image. At first glance, it seems that all the lines are curved, but in fact they are parallel. The illusion was discovered by R. Gregory at the Wall Cafe in Bristol. This is where its name came from.

Illusion of the Leaning Tower of Pisa

Above you see two pictures of the Leaning Tower of Pisa. At first glance, the tower on the right appears to lean more than the tower on the left, but in fact both of these pictures are the same. The reason is that the visual system views the two images as part of a single scene. Therefore, it seems to us that both photographs are not symmetrical.

Disappearing circles

This illusion is called "Vanishing Circles". It consists of 12 lilac pink spots arranged in a circle with a black cross in the middle. Each spot disappears in a circle for about 0.1 seconds, and if you focus on the central cross, you can get the following effect:
1) at first it will seem that there is a green spot running around
2) then the purple spots will start to disappear

Black and white illusion

Look at the four dots in the center of the picture for thirty seconds, then move your gaze to the ceiling and blink. What did you see?

fading

These are simple word problems from the Unified State Exam in Mathematics 2012. However, some of them are not so simple. For variety, some problems will be solved using Vieta’s theorem (see lesson “Vieta’s Theorem”), others - in a standard way, through a discriminant.

Of course, B12 problems will not always be reduced to a quadratic equation. Where a simple problem arises linear equation, no discriminants or Vieta’s theorems are required.

Task. For one of the monopolistic enterprises, the dependence of the volume of demand for products q (units per month) on its price p (thousand rubles) is given by the formula: q = 150 − 10p. Determine the maximum price level p (in thousand rubles), at which the value of the enterprise's revenue for the month r = q · p will be at least 440 thousand rubles.

This is a simple word problem. Let's substitute the demand formula q = 150 − 10p into the revenue formula r = q · p. We get: r = (150 − 10p) · p.

According to the condition, the company’s revenue must be at least 440 thousand rubles. Let's create and solve the equation:

(150 − 10p) p = 440 is quadratic equation;
150p − 10p 2 = 440 - opened the brackets;
150p − 10p 2 − 440 = 0 - collected everything in one direction;
p 2 − 15p + 44 = 0 - divided everything by the coefficient a = −10.

The result is the following quadratic equation. According to Vieta's theorem:
p 1 + p 2 = −(−15) = 15;
p 1 · p 2 = 44.

Obviously, the roots are: p 1 = 11; p2 = 4.

So, we have two candidates for the answer: the numbers 11 and 4. Let’s return to the problem statement and look at the question. It is required to find the maximum price level, i.e. from the numbers 11 and 4, you need to choose 11. Of course, this problem could also be solved through a discriminant - the answer would be exactly the same.

Task. For one of the monopolistic enterprises, the dependence of the volume of demand for products q (units per month) on their price p (thousand rubles) is given by the formula: q = 75 − 5p. Determine the maximum price level p (in thousand rubles), at which the value of the enterprise's revenue for the month r = q · p will be at least 270 thousand rubles.

The problem is solved similarly to the previous one. We are interested in revenue equal to 270. Since the enterprise’s revenue is calculated using the formula r = q · p, and demand is calculated using the formula q = 75 − 5p, let’s create and solve the equation:

(75 − 5p) p = 270;
75p − 5p 2 = 270;
−5p 2 + 75p − 270 = 0;
p 2 − 15p + 54 = 0.

The problem is reduced to the reduced quadratic equation. According to Vieta's theorem:
p 1 + p 2 = −(−15) = 15;
p 1 · p 2 = 54.

Obviously, the roots are the numbers 6 and 9. So, at a price of 6 or 9 thousand rubles, the revenue will be the required 270 thousand rubles. The problem asks you to indicate the maximum price, i.e. 9 thousand rubles.

Task. A model of a stone throwing machine shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y = ax 2 + bx, where a = −1/5000 (1/m), b = 1/10 are constant parameters. At what greatest distance (in meters) from a fortress wall 8 meters high should a machine be placed so that stones fly over it?

So, the height is given by the equation y = ax 2 + bx. In order for stones to fly over the fortress wall, the height must be greater or, in extreme cases, equal to the height of this wall. Thus, in the indicated equation the number y = 8 is known - this is the height of the wall. The remaining numbers are indicated directly in the condition, so we create the equation:

8 = (−1/5000) x 2 + (1/10) x - rather strong coefficients;
40,000 = −x 2 + 500x is already a completely sane equation;
x 2 − 500x + 40,000 = 0 - moved all terms to one side.

We obtained the reduced quadratic equation. According to Vieta's theorem:
x 1 + x 2 = −(−500) = 500 = 100 + 400;
x 1 x 2 = 40,000 = 100 400.

Roots: 100 and 400. We are interested in the greatest distance, so we choose the second root.

Task. A model of a stone throwing machine shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y = ax 2 + bx, where a = −1/8000 (1/m), b = 1/10 are constant parameters. At what greatest distance (in meters) from a 15-meter-high fortress wall should a machine be placed so that stones fly over it?

The task is completely similar to the previous one - only the numbers are different. We have:

15 = (−1/8000) x 2 + (1/10) x ;
120,000 = −x 2 + 800x - multiply both sides by 8000;
x 2 − 800x + 120,000 = 0 - collected all the elements on one side.

This is a reduced quadratic equation. According to Vieta's theorem:
x 1 + x 2 = −(−800) = 800 = 200 + 600;
x 1 x 2 = 120,000 = 200 600.

Hence the roots: 200 and 600. The largest root: 600.

Task. A model of a stone throwing machine shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y = ax 2 + bx, where a = −1/22,500 (1/m), b = 1/25 are constant parameters. At what greatest distance (in meters) from a fortress wall 8 meters high should a machine be placed so that stones fly over it?

Another problem with crazy odds. Height - 8 meters. This time we will try to solve through the discriminant. We have:

8 = (−1/22,500) x 2 + (1/25) x ;
180,000 = −x 2 + 900x - multiplied all numbers by 22,500;
x 2 − 900x + 180,000 = 0 - collected everything in one direction.

Discriminant: D = 900 2 − 4 · 1 · 180,000 = 90,000; Root of the discriminant: 300. Roots of the equation:
x 1 = (900 − 300) : 2 = 300;
x 2 = (900 + 300) : 2 = 600.

Largest root: 600.

Task. A model of a stone throwing machine shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y = ax 2 + bx, where a = −1/20,000 (1/m), b = 1/20 are constant parameters. At what greatest distance (in meters) from a fortress wall 8 meters high should a machine be placed so that stones fly over it?

Similar task. The height is again 8 meters. Let's create and solve the equation:

8 = (−1/20,000) x 2 + (1/20) x ;
160,000 = −x 2 + 1000x - multiply both sides by 20,000;
x 2 − 1000x + 160,000 = 0 - collected everything on one side.

Discriminant: D = 1000 2 − 4 1 160 000 = 360 000. Root of the discriminant: 600. Roots of the equation:
x 1 = (1000 − 600) : 2 = 200;
x 2 = (1000 + 600) : 2 = 800.

Largest root: 800.

Task. A model of a stone throwing machine shoots stones at a certain angle to the horizon with a fixed initial speed. Its design is such that the flight path of the stone is described by the formula y = ax 2 + bx, where a = −1/22,500 (1/m), b = 1/15 are constant parameters. At what greatest distance (in meters) from a 24-meter-high fortress wall should a machine be placed so that stones fly over it?

The next clone task. Required height: 24 meters. Let's make an equation:

24 = (−1/22,500) x 2 + (1/15) x ;
540,000 = −x 2 + 1500x - multiplied everything by 22,500;
x 2 − 1500x + 540,000 = 0 - collected everything in one direction.

We obtained the reduced quadratic equation. We solve using Vieta’s theorem:
x 1 + x 2 = −(−1500) = 1500 = 600 + 900;
x 1 x 2 = 540,000 = 600 900.

From the decomposition it is clear that the roots are: 600 and 900. We choose the largest: 900.

Task. A tap is fixed in the side wall of the cylindrical tank near the bottom. After opening it, water begins to flow out of the tank, and the height of the water column in it changes according to the law H (t) = 5 − 1.6t + 0.128t 2, where t is time in minutes. How long will it take for water to flow out of the tank?

Water will flow out of the tank as long as the height of the liquid column is greater than zero. Thus, we need to find out when H (t) = 0. We compose and solve the equation:

5 − 1.6t + 0.128t 2 = 0;
625 − 200t + 16t 2 = 0 - multiplied everything by 125;
16t 2 − 200t + 625 = 0 - arranged the terms in normal order.

Discriminant: D = 200 2 − 4 · 16 · 625 = 0. This means there will be only one root. Let's find it:

x 1 = (200 + 0) : (2 16) = 6.25. So, after 6.25 minutes the water level will drop to zero. This will be the moment until the water flows out.

Today's conversation is, to some extent, a continuation of the topic “Vertical text”. In addition to text written horizontally and vertically, we may need to write text, for example, at a certain angle, or even make it “lying” or tilted. We will talk about all this today.

The tool “Draw an inscription” will help us. Let’s open the “Insert” tab of the top menu and focus our attention on only the two functions it contains: “Shapes” and “Inscription”:

Both of these functionalities contain the same tool (option) “Draw an inscription”. Let’s expand on the contents of the “Shapes” functionality and see where the “Draw Label” tool is located:

So, the “Draw Lettering” tool is located in the “Basic Shapes” section of the shape set. If we once used this tool or some shape, then these shapes are reflected in the top section, with the name “Last used shapes”.

Now, without leaving the “Insert” tab, move the mouse cursor to its “Text” section and click the “Inscription” icon and in the window that opens, pay attention to the “Draw inscription” option:

This is still the same instrument. So, we have two options for activating the tool, no matter which way we go. Confirmation of the activity of the “Draw Label” tool will be a modification of the cursor - it will turn into a crosshair of two small lines:

By clicking and holding the left mouse button, we will create a field for text - draw a rectangle. The cursor will automatically be inside the rectangle, and we can start entering text:

So, the text entry is completed, you can start rotating it:

Last time, when we talked about “vertical text,” we rotated the text by grabbing the top green marker. Today we will act differently. I'll add two more lines of text to the box as an example.

The moment we finished drawing the field for the future text and released the left mouse button, significant changes occurred in the top menu. Completely independently (automatic mode), the options of the “Insert” tab were replaced by other options of the other “Format” tab:

But let's take a moment to rotate the text and pay attention to the field within which we place the text. The visibility of the field should not bother us, since we can make it invisible.

Why do we need to make the field invisible? And so that if text is written on a background with a color other than white, the working area of ​​the field is not visible.

So, let's make the field transparent using some of the options in the top menu's Format tab. Our task is to make the field truly transparent (now it is white) and remove its outline.

Let's start by removing the outline. To do this, expand the contents of the “Shape Outline” option and select the “No Outline” option from the list:

Now let’s make the field transparent, that is, reduce the white fill to zero. To do this, select the “Shape Fill” option and in the list of options that opens, select the “No fill” option:

This option may not always suit us, for the reason that “no fill” means the absence of a fill with a color other than white, as well as a gradient fill and a texture fill. That is, the field remained white as it was. In this particular case, this is an unnecessary action. Now I will place a triangle under the text, and we will make sure of this:

In order for the field to become truly transparent, we need to make other settings, and we will now make these same settings.

If the text field is not selected, then click in the text area to select it (the field is captured by markers). By left-clicking on the arrow in the lower right corner of the “Shape Styles” section of the “Format” tab, we will expand the additional settings window called “Shape Format”:

This window displays the settings that the field currently has. The field is filled with a solid white fill of 100% because the transparency level is 0%:

In order for the field to become completely transparent, we need to move the transparency slider to the right until a value equal to 100% appears in the window line. If we move the slider smoothly, we can observe how the text field becomes more and more transparent:

Having set the transparency level to 100%, click the “Close” button:

And here is the result of our actions:

Now let's move on to text rotation, as well as its tilt.

In order to rotate the text the way we want, we must, without leaving or collapsing the “Format” tab of the top menu, turn to the “Shape Effects” option:

And in the list of actions that opens, select the item “Rotate a volumetric figure”:

A new detailing window will open for us, where we will select the item “Rotation parameters for a volumetric figure”:

And now, finally, we get to the settings window:

In the lines where we currently see zero values ​​for the text rotation angles along the X, Y, Z axes, we set the desired values ​​by observing how the text rotates or tilts. We can set angles along all three coordinate axes, two or one. Or we can use the icons with blue arrows located in two columns to the right of the lines for entering numbers (tilt and rotation values). All we have to do is left-click on these very icons and look at what happens to the text:

In order to get into this window even faster, we need to left-click inside the text to select it, and then click the small arrow in the lower right corner of the “Shape Styles” section:

You should always first select text created using the Draw Text tool so that the required Drawing Tools Format tab appears in the top menu. And after it appears in the top menu, left-click on the name and expand the contents.

And this is the right window at our service:

And so that we can start setting parameters, we need to select the already familiar “Rotate volumetric figure” option:

We don’t necessarily have to enter the angle values ​​into any lines of the coordinate axes or click the icons with blue arrows to the right of the value entry lines. We can use the templates, a set of which is located at the top of the parameter settings window:

Let's left-click on the arrow-button to expand the list of blanks and select one or another blank, while simultaneously observing how the text behaves. I'll change the page orientation to landscape and increase the font size to make the changes easier to see:

By clicking the up and down arrows we can make the text in perspective:

If, for example, we set the X axis to 180 degrees, then our text will be “back to front”:

For additional influence on the text, in the same window we can use the “Inscription” option:

Well, in conclusion of today’s conversation about how to rotate text at an angle, as well as how to tilt text, I want to draw attention to important point. In order for us to twist the text like a pizzaiolo with dough, there should be no checkmark in the box labeled “Keep text flat”:

In geometry, an angle is a figure that is formed by two rays that emerge from one point (called the vertex of the angle). In most cases, the unit of measurement for angle is degree (°) - remember that a full angle, or one revolution, is 360°. You can find the angle value of a polygon by its type and the values ​​of other angles, and if given a right triangle, the angle can be calculated from two sides. Moreover, the angle can be measured using a protractor or calculated using a graphing calculator.

Steps

How to find interior angles of a polygon

Count the number of sides of the polygon. To calculate the interior angles of a polygon, you first need to determine how many sides the polygon has. Note that the number of sides of a polygon is equal to the number of its angles.

• For example, a triangle has 3 sides and 3 interior angles, and a square has 4 sides and 4 interior angles.

1. Calculate the sum of all interior angles of the polygon. To do this, use the following formula: (n - 2) x 180. In this formula, n is the number of sides of the polygon. The following are the sums of the angles of commonly encountered polygons:

   • The sum of the angles of a triangle (a polygon with 3 sides) is 180°.
   • The sum of the angles of a quadrilateral (a polygon with 4 sides) is 360°.
   • The sum of the angles of a pentagon (a polygon with 5 sides) is 540°.
   • The sum of the angles of a hexagon (a polygon with 6 sides) is 720°.
   • The sum of the angles of an octagon (a polygon with 8 sides) is 1080°.

2. Divide the sum of all the angles of a regular polygon by the number of angles. A regular polygon is a polygon with equal sides and equal angles. For example, each angle of an equilateral triangle is calculated as follows: 180 ÷ 3 = 60°, and each angle of a square is calculated as follows: 360 ÷ 4 = 90°.

   • An equilateral triangle and a square are regular polygons. And at the Pentagon building (Washington, USA) and road sign"Stop" shape of a regular octagon.

3. Subtract the sum of all known angles from the total sum of the angles of the irregular polygon. If the sides of a polygon are not equal to each other, and its angles are also not equal to each other, first add up the known angles of the polygon. Now subtract the resulting value from the sum of all the angles of the polygon - this way you will find the unknown angle.

   • For example, if given that the 4 angles of a pentagon are 80°, 100°, 120° and 140°, add up these numbers: 80 + 100 + 120 + 140 = 440. Now subtract this value from the sum of all the angles of the pentagon; this sum is equal to 540°: 540 - 440 = 100°. Thus, the unknown angle is 100°.

   Advice: the unknown angle of some polygons can be calculated if you know the properties of the figure. For example, in an isosceles triangle, two sides are equal and two angles are equal; In a parallelogram (which is a quadrilateral), opposite sides are equal and opposite angles are equal.

   Measure the length of the two sides of the triangle. Longest side right triangle called the hypotenuse. The adjacent side is the side that is near the unknown angle. The opposite side is the side that is opposite the unknown angle. Measure the two sides to calculate the unknown angles of the triangle.

   Advice: use a graphing calculator to solve the equations, or find an online table with the values ​​of sines, cosines, and tangents.

   Calculate the sine of an angle if you know the opposite side and the hypotenuse. To do this, plug the values ​​into the equation: sin(x) = opposite side ÷ hypotenuse. For example, the opposite side is 5 cm and the hypotenuse is 10 cm. Divide 5/10 = 0.5. Thus, sin(x) = 0.5, that is, x = sin -1 (0.5).

Let AB be some segment lying on a line, point M is an arbitrary point that does not belong to the line (Fig. 284). The angle a at the vertex M of the triangle AMB is called the angle at which the segment AB is visible from the point M. Let us find the locus of the points from which this segment is visible at the same constant angle a. To do this, we describe a circle around the triangle AMB and consider its arc AMB, containing point M. According to the previous one, from any point of the constructed arc, the segment AB will be visible at the same angle, measured by half of the arc ASB (in Fig. 284 it is shown by a dotted line). In addition, at the same angle the segment from will be visible. points of the arc located symmetrically with AMB relative to straight AB. From no other point of the plane, not lying on one of the found arcs, can the segment be visible at the same angle a.

In fact, from point P lying inside the figure bounded by the arcs AMB, the segment will be visible at an angle ARB greater than a, since the angle ARB will be measured by the half-sum of the arc ASB and some other arc, i.e. it will certainly be greater than the angle a. It is also clear that for an angle with vertex Q outside this figure we will have . Therefore, the points of the arcs AMB and AMB and only they have the required property: The geometric locus of the points from which a given segment is visible at a constant angle consists of two circular arcs symmetrically located relative to a given segment.

Problem 1. Given a segment AB and an angle a. Construct a segment containing the given angle a and resting on the segment AB. Here, a segment containing a given angle is understood as a segment bounded by a given segment and any of two circular arcs from the points of which the segment is visible at an angle a.

Solution. Let's draw a perpendicular to the segment AB in its middle (Fig. 285). The center of the circle whose segment needs to be constructed will be placed on this perpendicular. From end B of segment AB we draw a ray forming an angle with it; it will intersect the perpendicular at the center of the desired arc O (prove!).

Task 2. Construct a triangle using angle A, side and median.

Solution. On an arbitrary straight line we plot a segment BC equal to side a of the triangle (Fig. 286). The vertex of the triangle must be placed on the arc of the segment, from the points of which this segment is visible at angle a (the construction process is not shown in Fig. 286). Then from the middle M of side BC, as from the center, we draw a circle with a radius equal to m. The points of its intersection with the arc of the segment will give the possible positions of vertex A of the desired triangle. Explore the number of solutions!

Problem 3. Tangents to a circle are drawn from an external point. The tangent points divide the circle into parts whose ratio is equal to

Find the angle between the tangents.