Regarding arbitrary angle equalizers, it is best to write down the process of answering arbitrary an

2 Materials. Arbitrary equal angler.

It is not possible to divide any angle in 3 equal parts with a compass and a straightedge, but it can be done with the help of this equal angler. It was designed to divide arbitrary angles in 3 equal parts and then generalized to arbitrary equal angles.

The arbitrary anisoclator is made by nesting the bottom corners of isosceles triangles made of several transparencies in sequence. On each isosceles triangle, a groove is dug along the perpendicular line at the base, and the groove bisects the top angles of the isosceles triangle. After these isosceles triangles are sleeved in turn, as long as their common vertex c coincides with the vertex of the angle to be divided, and then the position of point c in the groove is moved, so as to adjust the angle of the vertex of each isosceles triangle, so that the first side and the last edge of any equal angle coincide with the two sides of the angle to be divided, and at this time, the line between the bottom corner vertex and the bottom edge vertical foot and the common vertex on any equal angle is exactly the arbitrary equal division of the angle to be divided.

The production data given here requires that the angle after the equalization should not be less than 9 degrees. If you want to divide it more finely, you need to reduce the top angle of each isosceles triangle appropriately. Examples of how to use it:

The trisect angle is shown in Fig. Fix 1 pin on the vertex O of a known angle pom, and select two pieces of film, (3 pieces of film for 5 equal parts, 4 pieces of film for 8 equal parts, and so on for the number of dots on the film); Put the narrow groove into the pin successively, move the film up and down, so that the first dot 1 on the left and dot 4 on the right just coincide on the two edges of the known angle POM, and then, insert the pencil into each of the dots to trace the marking points, remove the triangle film, write down and draw a line between the dots and the O dots in the POM, and complete the operation of dividing POM3 equally (the two round holes must coincide).

3×69°=23°；

As shown in the figure, let the ruler have a scale and pass the point A, let the grid line in the vertical direction of the side and the point B intersect at the point C, and the grid line in the horizontal direction of the point B intersects at the point D, keep the ruler with a scale on the side of the point A, adjust the position of the point C and D, so that CD=5cm, draw the ray AD, and MAD is what is sought

Method 1 (Cosine formula.

Starting from the corner vertices, the line segment a of the same length on the two sides is measured with a ruler, and then the distance b between the two endpoints is measured. Rule.

arccos[1-b^2/(2a^2)]

Method 2 (Making a ruler-type protractor).

If you haven't learned the cosine formula, you can only make a ruler-type protractor from a physical visual point of view. The method is to draw a semicircle with a known semicircular protractor, then align the right end of the ruler to the right side of the diameter, indicate 0° on the ruler with a pen, and then align the diameter, marking 180° at the intersection of the left side and the ring.

Then fix the right end of the support and do not move, clockwise.

Turn the left side, because the scale on the protractor is known, the scale at the intersection point can be faithfully reflected in 1° increments on the corresponding position of the ruler. Finally, add the ends and there are 181 scales. These scales are sparse on the left and dense on the right, and I wonder why I am doing this.

Finally, mark it at the midpoint of the ruler, which is equivalent to the corner vertex mark of the protractor, which should be used when measuring the angle.

When measuring any angle with this ruler protractor, first make the corner coincide with the corner vertex mark on the ruler, and extend each edge of the corner to the normal length of the ruler. Then fix 0° as the end of the right edge of the angle, rotate the ruler, when the ruler scale intersects with the left end of the angle, you can directly read the value on the ruler, and the scale between the estimates.

Make a right angle on one side and connect the two sides to form a right triangle. Then measure the ratio of the two sides at will, and use the trigonometric function to get the reading of the angle.

You can take one point on each of the two corner sides, connect them, measure the length of each side of the triangle, and use the cosine theorem to calculate the angle, cos = (a

This kind of topic also needs a process. You don't know much about this picture, but you can figure out what the hell you're going to do.

It is impossible to draw a ruler and a ruler to divide any angle into three equal parts. This is mathematically proven!

But there are many ways to use other tools, which are described here:

Archimedes' three-point method.

Plotting: 1 Set any acute angle AOB;

2 Take O as the center of the circle and make the circle O, AOB and the circle intersect at points A and B;

3 Extend the bo, to a considerable distance;

4 Intersect the ruler with the circle O, one point is A, and the other point is P;

5. At the same time, the extension line of the ruler and the BO intersect at point C;

6. Adjust the position of the ruler appropriately so that PC=AO;

7 with AC, then ACB=(1 3) AOB.

Proof: It can be proved by the relationship that the outer angles of the triangle are equal to the sum of the two internal angles that are not adjacent to each other; (omitted) explanation: Although this method does not conform to the formal ruler and gauge drawing, it provides a convenient and correct excellent means for the three points of the angle in practical work.

Make an arc with the vertices of the corners as the center of the circle, and intersect the edges of the corners at two points. Then take the two points as the center of the circle and make an arc (radius looks at it). The two arcs intersect a little. Connect the vertices of the point to the corner to get the corner bisector!

The principle is mainly derived from the fundamental properties of circles and the congruent triangle theorem.

Come on :-d

There is no problem that cannot be solved, as if it is not possible to make the length of the root number 2, and I think this method can also be ---

Make an arbitrary angle AOB, make an arc EF with its vertices, connect EF, and cross the fox EF to C and EF to DDraw a semicircle with d as the center of the circle and de as the radius. Intersect oc in h, divide the semicircle into three points (with e as the center of the circle, de as the radius to draw the fox on the semicircle, intersect with i, draw an arc with i as the center of the circle, call it in j), connect fi, intersect oc in k, make a perpendicular line perpendicular to oc with k as the vertical center, intersect fox ef as m, connect om, in m as the center of the circle, dm as the radius as the fox is called fox ef in n, connect on, you can pull.

Even the truth can be wrong, and even what people all over the world think is impossible can become possible. Respect the views of others is also respect for yourselves.

At the same time, this answer is my point of view, please respect my intellectual property rights and do not copy plagiarism.

Four lines can be made into three angles with a ruler, and the steps of Liang's three-point angle are shown in the figure.

Absolutely, it's just that people only know 1 3 = and ignore 3 = 1 + 2.

Don't think about it, impossible, since 1837 French mathematician.

Bai Wantzel proved the impossible with Galois theory. For those who claim that zhi uses rulers to make third-class dao angles, I ask you a few questions:1

Have you figured out what it means to draw a ruler? 2.Have you seen Vantzele's proof?

3.Have you read the proof from Wanzel? 4.

If you read Wanzel's proof, can you point out the error? 5.If you can't point out the error, then stop thinking about making thirds with a ruler.

Divided into three equal angles, ruler gauge drawing can be.

Divide the angle into thirds, draw the angle as a square.

Cut the triangle in half and turn it into a square. Calculate the area of the square, which can be drawn with a ruler and divided into three equal angles.

Solutionable. The chord segments are equal"method. Just math lovers understand!

It has long been proven that it is impossible to draw a ruler and a ruler to divide arbitrary angles into thirds.

Is it possible to draw only with a ruler and ruler, and to divide any angle into three equal parts?

Yes, now I'm struggling with how to publish it.

Trisected angles were one of the three major geometric problems of ancient Greece. The trisect angle is a famous problem in the ancient Greek geometric ruler diagram, and the problem of square and double cube is one of the three major problems of ancient mathematics, and now it has been proved that this problem is unsolvable. The full description of the issue is:

A given angle is divided into three equal parts using only a compass and an ungraduated ruler. Under the premise of ruler drawing (ruler drawing refers to drawing with a ruler and compass without scale), there is no solution to this problem. If the conditions are relaxed, such as allowing the use of graduated rulers, or if they can be used in conjunction with other curves, a given angle can be divided into thirds.

Once one of the world's famous problems (ruler and compass diagramming), using a ruler and compass to make an arbitrary angle of three equal parts, has long been proven impossible!