The sine theorem. Solving triangles

The study of triangles involuntarily raises the question of calculating the relationship between their sides and angles. In geometry, the cosine and sine theorem gives the most complete answer for solving this problem. In an abundance of different mathematical expressions and formulas, laws, theorems and rules, there are such that they differ in extraordinary harmony, conciseness and simplicity in conveying the meaning contained in them. The sine theorem is a vivid example of such a mathematical formulation. If in the verbal interpretation there is also a certain obstacle in understanding this mathematical rule, then when you look at the mathematical formula everything immediately falls into place.

The first information about this theorem was found in the form of its proof in the framework of the mathematical work of Nasir ad-Din Al-Tusi, dated the thirteenth century.

Approaching closer to the consideration of the aspect ratio and angles in any triangle, it is worth noting that the sine theorem allows solving a lot of mathematical problems, while this law of geometry finds its application in various types of practical human activity.

The sine theorem itself states that for any triangle the proportionality of the sides to the sines of opposite angles is characteristic. There is also the second part of this theorem, according to which the ratio of any side of the triangle to the sine of the opposite angle is equal to the diameter of the circle described near the triangle under consideration.

In the form of a formula, this expression looks like

A / sinA = b / sinB = c / sinC = 2R

Has a theorem of sine proof, which in various versions of textbooks is offered in a rich variety of versions.

For example, consider one of the proofs that explain the first part of the theorem. To this end, let us set the goal of proving the validity of the expression a SinC = C SinA.

In an arbitrary triangle ABC we construct the height BH. In one of the variants of the construction, H will lie on the segment AC, and in the other beyond its limits, depending on the angles at the vertices of the triangles. In the first case, the height can be expressed in terms of the angles and sides of the triangle, as BH = a sinC and BH = c sinA, which is the required proof.

In the case where the point H is outside the limits of the segment AC, we can obtain the following solutions:

BH = a sinC and BH = c sin (180-A) = c sinA;

Or BH = a sin (180-C) = a sinC and BH = c sinA.

As you can see, regardless of the construction options, we come to the desired result.

The proof of the second part of the theorem requires us to describe a circle around the triangle. Through one of the heights of the triangle, for example B, we construct the diameter of the circle. Obtain a point on the circle D with one of the height of the triangle, let it be point A of the triangle.

If we consider the resulting triangles ABD and ABC, then we can notice the equality of the angles C and D (they are based on one arc). And considering that the angle A is ninety degrees then sin D = c / 2R, or sin C = c / 2R, as required.

The sine theorem is the starting point for solving a wide range of different problems. A special attraction is its practical application, as a consequence of the theorem, we are able to relate the values of the sides of the triangle, the opposing angles and the radius (diameter) of the circle circumscribed around the triangle. The simplicity and accessibility of the formula describing this mathematical expression made it possible to widely use this theorem for solving problems using various mechanical counting devices (logarithmic rulers, tables, etc.), but even the arrival of powerful computing devices in the service of a man did not reduce the relevance of this theorem.

This theorem is not only included in the compulsory course of the geometry of the secondary school, but is further applied in certain branches of practical activity.