Does the sine theorem work in a right triangle. Sine theorem and cosine theorem. Variations and Generalizations

Sine theorem

The sine theorem establishes the relationship between the magnitude of the angles of a triangle and its opposite sides.

The formulation of the sine theorem:
The sides of a triangle are proportional to the sines of the opposite angles

Where
R- the radius of the circumscribed circle around the triangle
a, b, c- sides of a triangle
α, β, γ - the values of the angles opposite these sides

Proof of the sine theorem

The proof of the sine theorem occurs with the help of additional constructions.

Let's build an arbitrary triangle inscribed in a circle. Let's denote it as ABC.
Additionally, we construct the diameter of the circle, in which an arbitrary triangle is inscribed, but in such a way that it passes through one of its corners. 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's denote the diameter for the circumscribed circle as BD. The resulting triangle BCD is right-angled because its hypotenuse lies on the diameter of the circumscribed circle (a property of inscribed angles in a circle).

Thus, an additionally constructed triangle, which has one common side with an arbitrary triangle constructed earlier, and the hypotenuse coincides with the diameter of the circle - is rectangular. That is, the triangle DBC is a right triangle.

To prove the whole theorem, since the dimensions of the triangle ABC are chosen arbitrarily, it suffices to prove that the ratio of one arbitrary side to the angle opposite to it is equal to 2R.

Let it be 2R = a / sin α, that is, if we take according to the drawing 2R = BC / sin A.

Insofar as, angles inscribed in a circle, based on the same arc, are equal, then the angle CDB either equal to the angle CAB (if points A and D lie on the same side of line BC), or equal to π - CAB (otherwise).

Let's look at the properties trigonometric functions. Insofar as sin(π−α) = sinα, then the indicated options for constructing a triangle will still lead to the same result.

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 a right triangle.

2R=BC/sinA
2R=BC/(BC/DB)
2R=DB

And, since DB was built as the diameter of a circle, then the equality is true.
Repeating the same reasoning for the other two sides of the triangle, we get:

The sine theorem has been proven.

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 cosine and sine theorems 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 on the basis of the Pythagorean theorem: the square of the hypotenuse is equal to the sum squares of legs.

Consider an arbitrary triangle ABC. From the 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 the point C through trigonometric cos functions and sin.

Recall the definition of cosine and write 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.

Let's take the length AC as b and get the expression for the first coordinate of the point C:
x = b * cos⁡α. 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:

Obviously, both expressions (1) and (2) are equal to each other. We equate the right-hand sides and give similar ones:

In practice, this formula allows you to find the length of the unknown side of a triangle at given angles. The cosine theorem has three consequences: for a right, acute and obtuse angle of a triangle.

Let us replace the value of cos α with the usual variable x, then for the acute angle of the triangle ABC we get:

If the angle turns out to be right, then 2bx will disappear from the expression, since cos 90 ° \u003d 0. Graphically, the second consequence can be represented as follows:

In the case of an obtuse angle, the “-” sign in front of the double argument in the formula will change to “+”:

As you can see from the explanation, there is nothing complicated in the ratios. The cosine theorem is nothing more than an arrangement of the Pythagorean theorem in trigonometric quantities.

Practical application of the theorem

Exercise 1. Given a triangle ABC with side BC = a = 4 cm, AC = b = 5 cm, and cos α = ½. Find the length of side AB.

To correctly calculate, you need to determine the angle α. To do this, refer to the table of values for trigonometric functions, according to which the arc cosine is 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. It is required to find all the angles of the figure.

In this case, you can not do without a drawing of the conditions of the problem.

Since the angle values remain unknown, the full acute angle formula should be used to find solutions.

By analogy, it is not difficult to formulate and calculate the values of other angles:

In sum, the three angles of the triangle should be 180 °: 53 + 82 + 45 = 180, therefore, the solution is found.

Sine theorem

The theorem states that all sides of an arbitrary triangle are proportional to the sines of the opposite angles. The ratios are written in the form of a triple equality:

The classical proof of the statement is carried out on 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 also correspond to the sines of opposite angles, like 2R or D of a circle.

To do this, we draw the diameter of the circle from the vertex B. From the properties 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 a sine, the latter circumstance is not significant, since sin (π -α) \u003d sin α. Based on the above conclusions, it can be argued that:

sin ∠CGB = BC/ BG or sin A = BC/2R,

If we consider other angles of the figure, we get the extended formula of the sine theorem:

Typical tasks for practicing knowledge of the sine theorem come down to finding an unknown side or angle of a triangle.

As can be seen from the examples, the solution of such problems does not cause difficulties and consists in carrying out mathematical calculations.

Sine theorem- a theorem that establishes the dependence: the sides of a triangle are the angles opposite them.

Sine theorem: The sides of a triangle are proportional to the sines of the opposite angles.

There are 2 subspecies of the theorem: the ordinary and the extended sine theorem.

The usual sine theorem:

The sides of a triangle are proportional sin opposite corners.

Extended sine theorem for an arbitrary triangle:

where a, b, c- sides of a triangle , β, γ are the angles opposite these sides, and R is the radius of the circle that is circumscribed around the triangle.

Proof of the sine theorem.

Let there be a triangle inscribed in a circle. Let's denote it as ABC.

To prove the whole theorem, since the triangle has arbitrary dimensions, we can only prove that the ratio of the 1st arbitrary side to the opposite angle corresponds to 2R. Let's say it will 2R = a/sin, i.e. if you look at the drawing 2R=BC/sinA.

Let's draw a diameter | BG| for the circumscribed circle. From the properties of angles that are inscribed in a circle, the angle GCB will be straight, and the angle CGB equal to either when the points A And G are on the same side of the line BC, or − in the opposite way. Because sin(−)=sin, we get in both cases.

APPLYING THE THEOREM OF SINES

TASK 1

Consider in the section planimetry different types problems, the essence of the solution of which is reduced to the application of the sine theorem. Usually, at the beginning of solving such problems, actions of a transformative or simplifying nature are performed in order to bring the available data to a form that allows one to directly apply the sine theorem as the main tool for solving the problem. Examples of relevant solutions are needed for practical training in order to prepare well for a single state exam. The application of the sine theorem can be both the main action in solving a problem, and one of the necessary intermediate actions in solving more complex geometric problems. In this problem, the values of two angles of a triangle and one of the sides are known. You need to find the side of the triangle. Remember the solution! I wish you success!

In the triangle ABC . Find AC.