Stupid triangle. Stupid triangle let the corner with then side
1. Determine the type of triangle (acute, stupid or rectangular) with parties 8, 6 and 11 cm (Fig. 126). (one)
Decision. Denote the larger angle of the triangle through?. Obviously, he lies opposite the side of 11 cm, since the triangle larger angle lies against the main side. By the cosine theorem 112 \u003d 82+ 62- 2? 8? 6? COS?;
It was possible to argue differently. If angle? was equal to 90 °, then the big part of the Pythagore Theorem would be equal
The elongation of the side of 1 cm automatically increases and the angle under the face - it becomes blunt.
Answer: Stupid.
2. The base of the triangle is equal to 6 cm, one of the angles at the base is 105 °, the other is 45 °. Find the length of the side lying against an angle of 45 ° (Fig. 127). (one)
Decision. Suppose in the ABC triangle will be ac \u003d 6 cm ,? A \u003d 45 ° ,? C \u003d 105 °. Denote the length of the side of the sun through x. We need to find it. We use the sinus theorem on which:
Considering that the sum of the angles in the triangle is 180 °, we obtain :? B \u003d 180 ° -? A -? C \u003d 180 ° - 45 ° - 105 ° \u003d 30 °.
3. Find the triangle area with the parties 2,? 5 and 3 (Fig. 128). (one)
Decision. You can take advantage of the Gerona Formula:
In our case:
Semitter:
It would be easier to solve the task would be so. By the cosine theorem:
Since the triangle area is equal to half of the work of two sides on the sinus of the corner between them, then:
4. In the ABC triangle, where? ACB \u003d 120 °, a median was carried out. Find it length if the spear \u003d 6, Sun \u003d 4 (Fig. 129). (2)
Decision. We use the median length formula
We have a \u003d sun \u003d 4, b \u003d ac \u003d 6. It remains to find C \u003d AB. Apply to the triangle of the axle of cosine theorem: C2 \u003d AV2 \u003d AC2 + BC 2-2AC? BC? COS (? DC) \u003d 62+ 42- 2? 6? four ? COS 120 ° \u003d 36 + 16-48? (- 1/2) \u003d 76.
5. Find the lengths of the sides of the ABC of the ABC acute-angular triangle, if the sun \u003d 8, and the lengths of heights lowered on the AC and Sun side, are 6, 4 and 4, respectively (Fig. 130). (2)
Decision. The only triangle angle, which remained "intact", Corner C.
From the rectangular triangle of the Navy follows:
And now on the cosine theorem applied to the ABC triangle, we get:
Answer: AB \u003d? 41; AC \u003d 5.
6. In a triangle, one of the angles of which is equal to the difference between the other two, the length of the smaller side is equal to 1, and the sum of the squares of the squares built on two other sides, twice the area of \u200b\u200bthe area described near the triangle of the circle. Find the length of the larger side of the triangle (Fig. 131). (2)
Solution: Denote through? The smallest corner in the triangle and through? The greatest corner. Then the third corner is equal? -? -?. Under the condition of the task? -? \u003d? -? -? (A greater angle cannot be equal to the difference of two other corners). It follows that 2? \u003d?; ? \u003d? / 2. So the triangle is rectangular. Waste the aircraft lying against a smaller angle?, Equal under condition 1, which means that the second roll of Av is CTG?, And the AU hypotenuse is 1 / sin?. Therefore, the sum of the squares of the squares built on hypotenuse and larger nutta is:
The center of the circle described near the rectangular triangle lies in the middle of the hypotenuse, and its radius is equal to:
and the area is equal:
Using the condition of the task, we have an equation:
The length of the most side of the triangle is equal
7. The lengths of the side A, B, from the triangle are equal to 2, 3 and 4. Find the distance between the centers of the described and inscribed circles. (2)
Decision. To solve the problem, even the drawing is not needed. We consistently find: half-measure
Distance between centers of circles:
8. In the ABC triangle, the magnitude of the angle is equal to? / 3, the length of the height, lowered from the top with the side of the AB, is equal to? 3 cm, and the radius of the circle described near the ABC triangle is 5 cm. Find the lengths of the side of the ABC triangle (Fig. 132). (3)
Solution: Let the CD be the height of the ABC triangle, lowered from the summit C. Three cases are possible. The base D height CD gets:
1) on the segment AV;
2) to continue the segment of AV per point in;
3) to point V.
By the condition, the radius of the circle described near the ABC triangle is 5 cm. Consequently, in all three cases:
Now it is clear that the point D does not coincide with the point in, since the sun? CD. Using the Pythagora theorem to the triangles of ACD and BCD, we find that
It follows that point D lies between points A and B, but then Av \u003d AD + BD (1 + 6? 2), see
Answer: Av \u003d (6? 2 + 1) cm, sun \u003d 5? 3 cm, ac \u003d 2 cm.
9. In the triangles of ABC and A1B1C1, the length of the side AV is equal to the length of the side A1B1, the length of the speaker's side is equal to the length of the side A1C1, the angle value of you is 60 ° and the value of the angle B1A1C1 is 120 °. It is known that the ratio of length B1C1 to the length of the sun is equal to? N (where n is an integer). Find the ratio of the length of AU to the length of the AU. Under what values \u200b\u200bn task has at least one solution (Fig. 133)? (3)
Solution: Let ABC and A1B1C1 be the data on the triangle task condition. Applying cosine theorem to ABC and A1B1C1 triangles, we have:
T. K. under the condition of the task of B1C1: Sun \u003d? N, then
Since A1B1 \u003d AB and A1C1 \u003d AU, then, separating the numerator and denominator of the fraction on the left side of equality (1) on the AC2I, denoting AB: AU through x, we get equality:
where it is clear that the desired ratio of the length of AU to the length of the AS is the root of the equation
x2 (n - 1) - x (n + 1) + n - 1 \u003d 0. (2)
T. K. B1C1\u003e Sun, then n\u003e 1. Consequently, equation (2) is square. Its discriminant is equal to (n + 1) 2- 4 (n - 1) 2 \u003d - 3N2 + 10N - 3.
Equation (2) will have solutions if - 3n2 + 10n - 3? 0, i.e. at -1/3? n? 3. T. K. N is a natural number, greater than 1, then equation (2) has solutions at n \u003d 2 and n \u003d 3. With n \u003d 3, equation (2) has a root x \u003d 1; For n \u003d 2, the equation has a root
Answer: The ratio of the length of AB to the length of the speaker is equal
at n \u003d 2; equal to 1 at n \u003d 3; With the remaining N solutions there.
In general, the triangle is the most simplest figure of all existing polygons. It is formed with the help of three points, which lie in the 1st plane, but, at the same time they do not lie on the 1st straight, and pairs are connected to each other. Triangles are of different types, and therefore are characterized by different properties. Depending on the type of angles, the triangle may relate to one of the 3 types - to be acutely angular, rectangular or stupid. The stupid triangle is a triangle that has one stupid angle. At the same time, stupid is called such an angle, which has the magnitude of more ninety degrees, but less than one hundred eighty degrees.
In other words, a stupid triangle is the simplest polygon, which contains a stupid angle - some of its corners are within 90-180 degrees.
Task: Is there or not a triangle stupid when:
- aBC angle in it equals 65 degrees;
- its BCA angle is 95 degrees;
- cAB angle - 20 degrees.
Solution: CAB and ABC corners are less than 90 degrees, but, with the BCA angle of more than 90 degrees. So, such a triangle is stupid.
How to find the sides of a stupid anose-free triangle
What is a stupid triangle, we dealt above. Now it should be dealt with which triangle is considered an equally chaired.
Equally called such a triangle, which has 2 absolutely equal side. These sides are called side, the third side of the triangle is called the basis.
The vertices of the triangle are usually indicated by capital latin letters - that is, A, B and C. The values \u200b\u200bof its corners, respectively, the vertices are designated by Greek letters, that is, α, β, γ. The lengths of the opposite sides of the triangle are capital latin letters, that is, a, b, c.
A simple task: the perimeter of a stupid iscessed triangle is 25cm, the difference of 2 of its sides is 4 cm, and the 1-in from the external corners of the triangle is sharp. How to find such a triangle?
Solution: an angle adjacent to which the acute angle of the triangle is stupid. In a triangle of such a plan, a blunt angle may be exclusively the angle that is against its foundation. Accordingly, the base is the largest side of such a triangle. If you take the base of this triangle for x, then to solve this problem you need to use the following formula:
Answer: The basis of an equally chained stupid triangle is 11 cm, and its both sides of 7 cm.
Formulas for which you can find the sides of a stupid anose-free triangle
Notation used:
- b - this is the side of the base of the triangle
- a - His equal side
- α - angles at the base of the triangle
- β - an angle that is formed by its equal parties
- √ - square root
1. Formulas of the base length (b):
- b \u003d 2a sin (β / 2) \u003d A√2-2COSβ
- b \u003d 2a COS α
2. Formulas of the length of the equal sides of the triangle (s):
2Sin (β / 2) √2-2cos β
How to find a cosine angle in a stupid triangle if the height is known
To begin, it will not hurt to understand with the main terms that are used in this matter: what is called the height of the triangle and what is cosine angle.
The height of the triangle is considered perpendicular, which is carried out from the top of it to the line, which contains the opposite side of this triangle. Cosine is a well-known trigonometric function, which is one of the main functions of trigonometry.
In order to find the cosine of the angle in a stupid triangle with the vertices A, B and C, provided that the height is known, you need to lower the height from the side of the speakers. The point in which the height intersects with a side of the AU must be denoted by D and consider the triangle of the AVD, which is rectangular. In this triangle AB, which is a side of the original triangle, is hypotenuse. Catests are the height of the original triangle, as well as the segment of the AD, which belongs to the side of the AU. At the same time, the cosine of the angle corresponding to the top A is equal to the attitude of the AD to AB, since the AD catat is adjacent to the corner at the top of the AV triangle. In the case when it is known what exactly the ratio of the AU share is divided by the VD height and what is this height, then the cosine of the angle corresponding to the vertex A, found.
Question 1.What angles are called adjacent?
Answer.Two angles are called adjacent if they have one side in common, and other parties of these angles are additional semicircles.
In Figure 31, the angles (A 1 B) and (A 2 B) adjacent. They have the side B overall, and the parties A 1 and A 2 are additional semicircles.
Question 2.Prove that the sum of adjacent angles is 180 °.
Answer. Theorem 2.1.The sum of adjacent angles is 180 °.
Evidence. Let an angle (A 1 b) and an angle (A 2 B) - these adjacent angles (see Fig.31). Beam B passes between the sides of A 1 and A 2 of the deployed corner. Therefore, the sum of the angles (A 1 b) and (A 2 B) is equal to the deployed corner, i.e. 180 °. Q.E.D.
Question 3.Prove that if two angles are equal, then adjacent angles are also equal.
Answer.
From Theorem 2.1
it follows that if two angles are equal, then the adjacent angles are equal.
Suppose the angles (A 1 B) and (C 1 D) are equal. We need to prove that the angles (A 2 B) and (C 2 D) are also equal.
The sum of adjacent angles is 180 °. It follows from this that A 1 B + A 2 B \u003d 180 ° and C 1 D + C 2 D \u003d 180 °. Hence, A 2 B \u003d 180 ° - A 1 B and C 2 D \u003d 180 ° - C 1 D. Since the angles (A 1 b) and (C 1 D) are equal, we obtain that A 2 B \u003d 180 ° - A 1 B \u003d C 2 D. According to the transitivity property of the equality sign, it follows that A 2 B \u003d C 2 D. Q.E.D.
Question 4.What angle is called direct (sharp, stupid)?
Answer. An angle equal to 90 ° is called a direct angle.
An angle less than 90 ° is called a sharp angle.
The angle greater than 90 ° and the smaller 180 ° is called stupid.
Question 5. Prove that angle, adjacent to direct, is a straight angle.
Answer.From the theorem on the sum of adjacent angles it follows that the angle, adjacent to the direct angle, is a direct angle: x + 90 ° \u003d 180 °, x \u003d 180 ° - 90 °, x \u003d 90 °.
Question 6.What angles are called vertical?
Answer.Two angles are called vertical if the sides of the same angle are additional semi-simply sides of the other.
Question 7.Prove that vertical angles are equal.
Answer. Theorem 2.2. Vertical angles are equal.
Evidence.Let (A 1 B 1) and (A 2 B 2) - these vertical angles (Fig. 34). The angle (A 1 B 2) is adjacent with an angle (A 1 B 1) and with an angle (A 2 B 2). Hence the theorem on the sum of adjacent angles, we conclude that each of the angles (A 1 B 1) and (A 2 B 2) complements the angle (A 1 B 2) to 180 °, i.e. Angles (A 1 B 1) and (A 2 B 2) are equal. Q.E.D.
Question 8.Prove that if with the intersection of two straight lines one of the corners of the line, then the remaining three angle is also straight.
Answer.Suppose that direct AB and CD crosses each other at the point O. Suppose the AOD angle is 90 °. Since the sum of adjacent angles is 180 °, we obtain that AOC \u003d 180 ° -aod \u003d 180 ° is 90 ° \u003d 90 °. Cob angle vertical AOD angle, so they are equal. That is, the angle cob \u003d 90 °. COA angle vertical corner BOD, so they are equal. That is, the angle bod \u003d 90 °. Thus, all the angles are 90 °, that is, they are all direct. Q.E.D.
Question 9.What are the direct are called perpendicular? What sign is used to refer to the perpendicularity of direct?
Answer.Two straight lines are called perpendicular if they intersect at right angles.
The perpendicularity of the direct is denoted by the sign \\ (\\ perp \\). Record \\ (A \\ PERP B \\) reads: "Direct A perpendicular to direct b".
Question 10.Prove that through any point the straight can be carried out by the person perpendicular to it, and only one.
Answer. Theorem 2.3.Through each direct can be carried out directly, and only one.
Evidence.Let A be this direct and a - this point on it. Denote by a 1 one of the semiconductible direct A with the starting point A (Fig. 38). We will postpone from the semicircular A 1 angle (A 1 B 1), equal to 90 °. Then the direct containing the beam B 1 will be perpendicular to the direct a.
Suppose that there is another straight line, also passing through the point A and perpendicular to the straight line a. Denote by C 1, the semi-axis of this straight line, lying in one half-plane with a beam B 1.
The angles (A 1 B 1) and (A 1 C 1), equal to each 90 °, are postponed in one half-plane from the semi-simplicable A 1. But from the semiconduma a 1 in this half-plane, only one angle can be postponed equal to 90 °. Therefore, not to be another direct passing through the point A and perpendicular direct a. Theorem is proved.
Question 11.What is perpendicular to the straight line?
Answer. The perpendicular to this direct is called a straight line, perpendicular to this, which has one of its ends their intersection point. This end of the segment is called base Perpendicular.
Question 12.Explain that the proof of nasty.
Answer. The method of evidence that we applied in Theorem 2.3 is called proof of the opponent. This method of evidence is that we initially make an assumption that is the opposite of what is approved by the theorem. Then, by reasoning, relying on axioms and proven theorems, come to the conclusion that is contrary to either the condition of the theorem or one of the axioms or a previously proven theorem. On this basis, we conclude that our assumption was incorrect, and therefore the statement of the theorem is true.
Question 13.What is called bisector angle?
Answer.The bisector of the angle is called a beam, which comes from the top of the corner, passes between its parties and divides the angle in half.
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