Theorem of sines
The sine theorem establishes the relationship between the size of the angles of a triangle and its opposite sides.Statement of the sine theorem:
The sides of a triangle are proportional to the sines of the opposite angles
Where
R- radius of the circle circumscribed around the triangle
a, b, c- sides of a triangle
α, β, γ
- the magnitude of the angles opposite to these sides
Proof of the Theorem of Sines
The proof of the theorem of sines occurs using additional constructions.
Let's construct an arbitrary triangle, inscribed in a circle. Let's denote it as ABC.
Additionally, we construct the diameter of the circle, into which an arbitrary triangle is inscribed, but so that it passes through one of its angles. The diameter is equal to twice the radius of the circle (2R).
Let us take into account that one of the properties of a right triangle inscribed in a circle is that its hypotenuse is the diameter of the circle in which it is inscribed.
Let us denote the diameter of the circumcircle as BD. The resulting triangle BCD is right-angled because its hypotenuse lies on the diameter of the circumscribed circle (the property of angles inscribed in a circle).
Thus, an additionally constructed triangle, which has one common side with the previously constructed arbitrary triangle, and the hypotenuse coincides with the diameter of the circle - is rectangular. That is, the triangle DBC is right-angled.
To prove the entire theorem, since the dimensions of triangle ABC are chosen arbitrarily, it is enough to prove that the ratio of one arbitrary side to the angle opposite it is equal to 2R.
Let it be 2R = a / sin α, that is, if we take from the drawing 2R = BC / sin A.
Because the, Angles inscribed in a circle and subtending the same arc are equal, then angle CDB is either equal to angle CAB (if points A and D lie on the same side of line BC), or equal to π - CAB (otherwise).
Let us turn to the properties of trigonometric functions. Because the sin(π − α) = sin α, then the indicated options for constructing a triangle will still lead to the same result.
Let's calculate the value 2R = a / sin α, according to the drawing 2R = BC / sin A. To do this, replace sin A with the ratio of the corresponding sides of the right triangle.
2R = BC / sin A
2R = BC / (BC / DB)
2R = DB
And, since DB was constructed as the diameter of a circle, then the equality is satisfied.
Repeating the same reasoning for the other two sides of the triangle, we get:
The sine theorem has been proven.
The sides of a triangle are proportional to the sines of the opposite angles.
Proof:
Let triangle ABC, side AB = c, side BC = a, side CA = b.
Let's try to prove that a/sin(A) = b/sin(B) = c/sin(C). Let's use the triangle area theorem and write it for each pair of sides and their corresponding angle:
S = (1/2)*a*b*sin(C),
S = (1/2)*b*c*sin(A),
S = (1/2)*c*a*sin(B).
Since the left sides of the first two equalities are the same, the right sides can be equated with each other. We get (1/2)*a*b*sin(C) = (1/2)*b*c*sin(A). Let's reduce this equality by ½*b, we get:
a*sin(C) = c*sin(A).
a/sin(A) = c/sin(C).
Since the left parts of the second and third equalities are the same, the right parts can be equated with each other. We get (1/2)*b*c*sin(C) = (1/2)*c*a*sin(B). Let's reduce this equality by 1/2*c, we get:
b*sin(A) = a*sin(B).
By the property of proportion we get:
a/sin(A) = b/sin(B).
Combining the two results we get: a/sin(A) = b/sin(B) = c/sin(C). Q.E.D.
The solution of the problem
The following fact can also be proven. The ratio of any side of a triangle to the sine of the opposite angle is equal to the diameter of the circumscribed circle of the triangle.
In other words, for any triangle ABC with side AB = c, side BC = a, side CA = b, the following equalities hold: a/sin(A) = b/sin(B) = c/sin(C) = 2*R. Here R is the radius of the circle circumscribed about the triangle.
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Trigonometry is widely used not only in the section of algebra - the beginning of analysis, but also in geometry. In this regard, it is reasonable to assume the existence of theorems and their proofs related to trigonometric functions. Indeed, the theorems of cosines and sines derive very interesting, and most importantly useful, relationships between the sides and angles of triangles.
Using this formula, you can derive any of the sides of the triangle:
The proof of the statement is derived based on the Pythagorean theorem: the square of the hypotenuse is equal to the sum of the squares of the legs.
Consider an arbitrary triangle ABC. From vertex C we lower the height h to the base of the figure; in this case its length is absolutely not important. Now, if we consider an arbitrary triangle ACB, then we can express the coordinates of point C through the trigonometric functions cos and sin.
Let's remember the definition of cosine and write down the ratio of the sides of the triangle ACD: cos α = AD/AC | multiply both sides of the equality by AC; AD = AC * cos α.
We take the length AC as b and obtain an expression for the first coordinate of point C:
x = b * cosα. Similarly, we find the value of the ordinate C: y = b * sin α. Next, we apply the Pythagorean theorem and express h alternately for the triangle ACD and DCB:
It is obvious that both expressions (1) and (2) are equal to each other. Let’s equate the right-hand sides and present similar ones:
In practice, this formula allows you to find the length of the unknown side of a triangle from given angles. The cosine theorem has three consequences: for right, acute and obtuse angles of a triangle.
Let us replace the value of cos α with the usual variable x, then for the acute angle of triangle ABC we obtain:
If the angle turns out to be right, then 2bx will disappear from the expression, since cos 90° = 0. Graphically, the second consequence can be represented as follows:
In the case of an obtuse angle, the “-” sign before the double argument in the formula will change to “+”:
As can be seen from the explanation, there is nothing complicated in the relationships. The cosine theorem is nothing more than a translation of the Pythagorean theorem into trigonometric quantities.
Practical application of the theorem
Exercise 1. Given a triangle ABC, whose side BC = a = 4 cm, AC = b = 5 cm, and cos α = ½. You need to find the length of side AB.
To make the calculation correctly, you need to determine the angle α. To do this, you should refer to the table of values for trigonometric functions, according to which the arc cosine is equal to 1/2 for an angle of 60°. Based on this, we use the formula of the first corollary of the theorem:
Task 2. For triangle ABC, all sides are known: AB =4√2,BC=5,AC=7. You need to find all the angles of the figure.
In this case, you cannot do without a drawing of the conditions of the problem.
Since the angle values remain unknown, the full formula for an acute angle should be used to find solutions.
By analogy, it is not difficult to create formulas and calculate the values of other angles:
The sum of the three angles of the triangle should be 180°: 53 + 82 + 45 = 180, therefore, the solution has been found.
Theorem of sines
The theorem states that all sides of an arbitrary triangle are proportional to the sines of the opposite angles. The relations are written in the form of triple equality:
The classical proof of the statement is carried out using the example of a figure inscribed in a circle.
To verify the veracity of the statement using the example of triangle ABC in the figure, it is necessary to confirm the fact that 2R = BC / sin A. Then prove that the other sides are related to the sines of opposite angles, like 2R or D of a circle.
To do this, draw the diameter of the circle from vertex B. From the property of angles inscribed in a circle, ∠GCB is a straight line, and ∠CGB is either equal to ∠CAB or (π - ∠CAB). In the case of sine, the latter circumstance is not significant, since sin (π –α) = sin α. Based on the above conclusions, it can be stated that:
sin ∠CGB = BC/ BG or sin A = BC/2R,
If we consider other angles of the figure, we obtain an extended formula for the theorem of sines:
Typical tasks for practicing the sine theorem boil down to finding an unknown side or angle of a triangle.
As can be seen from the examples, solving such problems is not difficult and consists of carrying out mathematical calculations.
Let's construct an arbitrary triangle inscribed in a circle. Let's denote it as ABC.
To prove the entire theorem, since the dimensions of the triangle are chosen arbitrarily, it is enough to prove that the ratio of one arbitrary side to the angle opposite it is equal to 2R. Let it be 2R = a / sin α, that is, if we take from the drawing 2R = BC / sin A.
Let us calculate the diameter BD for the circumscribed circle. The resulting triangle BCD is right-angled because its hypotenuse lies on the diameter of the circumscribed circle (the property of angles inscribed in a circle).
Since the angles inscribed in a circle and resting on the same arc are equal, then angle CDB is either equal to angle CAB (if points A and D lie on the same side of line BC), or equal to π - CAB (otherwise) .
Let us turn to the properties of trigonometric functions. Since sin(π − α) = sin α, the indicated options for constructing a triangle will still lead to the same result.
Let's calculate the value 2R = a / sin α, according to the drawing 2R = BC / sin A. To do this, replace sin A with the ratio of the corresponding sides of the right triangle.
2R = BC / sin A
2R = BC / (BC / DB)
2R = DB
And, since DB was constructed as the diameter of a circle, then the equality is satisfied.
Repeating the same reasoning for the other two sides of the triangle, we get: 
The sine theorem has been proven.
Theorem of sines
Note. This is part of a lesson with geometry problems (section theorem of sines). If you need to solve a geometry problem that is not here, write about it in the forum. In tasks, instead of the "square root" symbol, the sqrt() function is used, in which sqrt is the square root symbol, and the radical expression is indicated in brackets.
Theorem of sines:
The sides of a triangle are proportional to the sines of the opposite angles, or, in an expanded formulation:
a / sin α = b / sin β = c / sin γ = 2R
where R is the radius of the circumscribed circle
For the theory - the formulation and proof of the theorem, see in detail in the chapter "Theorem of Sines" .
Task
In triangle XYZ, angle X=30, angle Z=15. The perpendicular YQ to ZY divides the side XZ into parts XQ and QZ. Find XY if QZ = 1.5 m
Solution.
The height formed two right triangles XYQ and ZYQ.
To solve the problem, we will use the theorem of sines.
QZ / sin(QYZ) = QY / sin(QZY)
QZY = 15 degrees, Accordingly, QYZ = 180 - 90 - 15 = 75
Since the length of the altitude of the triangle is now known, let's find XY using the same theorem of sines.
QY / sin(30) = XY / sin(90)
Let's take into account the tabular values of some trigonometric functions:
- sine of 30 degrees is equal to sin(30) = 1 / 2
- the sine of 90 degrees is equal to sin(90) = 1
QY = XY sin (30)
3/2 (√3 - 1) / (√3 + 1) = 1/2 XY
XY = 3 (√3 - 1) / (√3 + 1) ≈ 0.8 m
Answer: 0.8 m or 3 (√3 - 1) / (√3 + 1)
Theorem of sines (part 2)
Note. This is part of a lesson with geometry problems (section theorem of sines). If you need to solve a geometry problem that is not here, write about it in the forum .
See the theory in detail in the chapter "Theorem of Sines" .
Task
Side AB of triangle ABC is 16cm. Angle A is 30 degrees. Angle B is 105 degrees. Calculate the length of side BC. 
Solution.
According to the law of sines, the sides of a triangle are proportional to the sines of the opposite angles:
a / sin α = b / sin β = c / sin γ
Thus
BC / sin α = AB / sin γ
We find the size of angle C based on the fact that the sum of the angles of a triangle is equal to 180 degrees.
C = 180 - 30 -105 = 45 degrees.
Where:
BC / sin 30° = 16 / sin 45°
BC = 16 sin 30° / sin 45°
Referring to the table of trigonometric functions, we find:
BC = (16 * 1 / 2) / √2/2 = 16 / √2 ≈ 11.3 cm
Answer: 16 / √2
Task.
In triangle ABC, angle A = α, angle C = β, BC = 7cm, BN is the height of the triangle.
Find AN 
