Formula for finding the area of a right triangle using sine. Games with ruler and pencil
In life, we will often have to deal with mathematical problems: at school, at university, and then helping our child with completing homework. People in certain professions will encounter mathematics on a daily basis. Therefore, it is useful to memorize or recall mathematical rules. In this article we will analyze one of them: finding the leg right triangle.
What is a right triangle
First, let's remember what a right triangle is. A right triangle is geometric figure of three segments that connect points that do not lie on the same straight line, and one of the angles of this figure is 90 degrees. The sides forming a right angle are called legs, and the side that lies opposite right angle– hypotenuse.
Finding the leg of a right triangle
There are several ways to find out the length of the leg. I would like to consider them in more detail.
Pythagorean theorem to find the side of a right triangle
If we know the hypotenuse and the leg, then we can find the length of the unknown leg using the Pythagorean theorem. It sounds like this: “Square of the hypotenuse equal to the sum squares of legs.” Formula: c²=a²+b², where c is the hypotenuse, a and b are the legs. We transform the formula and get: a²=c²-b².
Example. The hypotenuse is 5 cm, and the leg is 3 cm. We transform the formula: c²=a²+b² → a²=c²-b². Next we solve: a²=5²-3²; a²=25-9; a²=16; a=√16; a=4 (cm).
Trigonometric ratios to find the leg of a right triangle
You can also find an unknown leg if any other side and any acute angle of a right triangle are known. There are four options for finding the leg using trigonometric functions: by sine, cosine, tangent, cotangent. The table below will help us solve problems. Let's consider these options.
Find the leg of a right triangle using sine
The sine of an angle (sin) is the ratio of the opposite side to the hypotenuse. Formula: sin=a/c, where a is the leg opposite the given angle, and c is the hypotenuse. Next, we transform the formula and get: a=sin*c.
Example. The hypotenuse is 10 cm, angle A is 30 degrees. Using the table, we calculate the sine of angle A, it is equal to 1/2. Then, using the transformed formula, we solve: a=sin∠A*c; a=1/2*10; a=5 (cm).
Find the leg of a right triangle using cosine
The cosine of an angle (cos) is the ratio of the adjacent leg to the hypotenuse. Formula: cos=b/c, where b is the leg adjacent to a given angle, and c is the hypotenuse. Let's transform the formula and get: b=cos*c.
Example. Angle A is equal to 60 degrees, the hypotenuse is equal to 10 cm. Using the table, we calculate the cosine of angle A, it is equal to 1/2. Next we solve: b=cos∠A*c; b=1/2*10, b=5 (cm).
Find the leg of a right triangle using tangent
Tangent of an angle (tg) is the ratio of the opposite side to the adjacent side. Formula: tg=a/b, where a is the side opposite to the angle, and b is the adjacent side. Let's transform the formula and get: a=tg*b.
Example. Angle A is equal to 45 degrees, the hypotenuse is equal to 10 cm. Using the table, we calculate the tangent of angle A, it is equal to Solve: a=tg∠A*b; a=1*10; a=10 (cm).
Find the leg of a right triangle using cotangent
Angle cotangent (ctg) is the ratio of the adjacent side to the opposite side. Formula: ctg=b/a, where b is the leg adjacent to the angle, and is the opposite leg. In other words, cotangent is an “inverted tangent.” We get: b=ctg*a.
Example. Angle A is 30 degrees, the opposite leg is 5 cm. According to the table, the tangent of angle A is √3. We calculate: b=ctg∠A*a; b=√3*5; b=5√3 (cm).
So now you know how to find a leg in a right triangle. As you can see, it’s not that difficult, the main thing is to remember the formulas.
Sine is one of the basic trigonometric functions, the use of which is not limited to geometry alone. Tables for calculating trigonometric functions, like engineering calculators, is not always at hand, and calculating the sine is sometimes necessary to solve various tasks. In general, calculating the sine will help consolidate drawing skills and knowledge of trigonometric identities.
Games with ruler and pencil
A simple task: how to find the sine of an angle drawn on paper? To solve, you will need a regular ruler, a triangle (or compass) and a pencil. The simplest way to calculate the sine of an angle is by dividing the far leg of a triangle with a right angle by the long side - the hypotenuse. Thus, you first need to complete the acute angle to the shape of a right triangle by drawing a line perpendicular to one of the rays at an arbitrary distance from the vertex of the angle. We will need to maintain an angle of exactly 90°, for which we need a clerical triangle.
Using a compass is a little more accurate, but will take more time. On one of the rays you need to mark 2 points at a certain distance, set a radius on the compass approximately equal to the distance between the points, and draw semicircles with centers at these points until the intersections of these lines are obtained. By connecting the intersection points of our circles with each other, we get a strict perpendicular to the ray of our angle; all that remains is to extend the line until it intersects with another ray.
In the resulting triangle, you need to use a ruler to measure the side opposite the corner and the long side on one of the rays. The ratio of the first dimension to the second will be the desired value of the sine acute angle.
Find the sine for an angle greater than 90°
For an obtuse angle the task is not much more difficult. We need to draw a ray from the vertex in the opposite direction using a ruler to form a straight line with one of the rays of the angle we are interested in. The resulting acute angle should be treated as described above; the sines of adjacent angles that together form a reverse angle of 180° are equal.
Calculating sine using other trigonometric functions
Also, calculating the sine is possible if the values of other trigonometric functions of the angle or at least the lengths of the sides of the triangle are known. Trigonometric identities will help us with this. Let's look at common examples.
How to find the sine with a known cosine of an angle? The first trigonometric identity, based on the Pythagorean theorem, states that the sum of the squares of the sine and cosine of the same angle is equal to one.
How to find the sine with a known tangent of an angle? The tangent is obtained by dividing the far side by the near side or dividing the sine by the cosine. Thus, the sine will be the product of the cosine and the tangent, and the square of the sine will be the square of this product. We replace the squared cosine with the difference between unity and the square sine according to the first trigonometric identity and, through simple manipulations, we reduce the equation to the calculation of the square sine through the tangent; accordingly, to calculate the sine, you will have to extract the root of the result obtained.
How to find the sine with a known cotangent of an angle? The value of the cotangent can be calculated by dividing the length of the leg closest to the angle by the length of the far one, as well as dividing the cosine by the sine, that is, the cotangent is a function inverse to the tangent relative to the number 1. To calculate the sine, you can calculate the tangent using the formula tg α = 1 / ctg α and use the formula in the second option. You can also derive a direct formula by analogy with tangent, which will look like this.
How to find the sine of three sides of a triangle
There is a formula for finding the length of the unknown side of any triangle, not just a right triangle, from two known sides using the trigonometric function of the cosine of the opposite angle. She looks like this.
The area of a triangle is equal to half the product of its sides and the sine of the angle between them.
Proof:
Consider an arbitrary triangle ABC. Let side BC = a, side CA = b and S be the area of this triangle. It is necessary to prove that S = (1/2)*a*b*sin(C).
To begin with, let's introduce a rectangular coordinate system and place the origin of coordinates at point C. Let's position our coordinate system so that point B lies on the positive direction of the Cx axis, and point A has a positive ordinate.
If everything is done correctly, you should get the following drawing.
Square given triangle can be calculated using the following formula: S = (1/2)*a*h, where h is the height of the triangle. In our case, the height of the triangle h is equal to the ordinate of point A, that is, h = b*sin(C).
Taking into account the results obtained, the formula for the area of a triangle can be rewritten as follows: S = (1/2)*a*b*sin(C). Q.E.D.
Problem solving
Problem 1. Find the area triangle ABC, if a) AB = 6*√8 cm, AC = 4 cm, angle A = 60 degrees b) BC = 3 cm, AB = 18*√2 cm, angle B = 45 degrees c) AC = 14 cm, CB = 7 cm, angle C = 48 degrees.
According to the theorem proved above, the area S of triangle ABC is equal to:
S = (1/2)*AB*AC*sin(A).
Let's do the calculations:
a) S = ((1/2) *6*√8*4*sin(60˚)) = 12*√6 cm^2.
b) S = (1/2)*BC*BA*sin(B)=((1/2)* 3*18*√2 *(√2/2)) = 27 cm^2.
c) S = (1/2)*CA*CB*sin(C) = ½*14*7*sin48˚ cm^2.
We calculate the value of the sine of an angle on a calculator or use the values from the table of values of trigonometric angles. Answer:
a) 12*√6 cm^2.
c) approximately 36.41 cm^2.
Problem 2. The area of triangle ABC is 60 cm^2. Find side AB if AC = 15 cm, angle A = 30˚.
Let S be the area of triangle ABC. By the theorem on the area of a triangle we have:
S = (1/2)*AB*AC*sin(A).
Let’s substitute the values we have into it:
60 = (1/2)*AB*15*sin30˚ = (1/2)*15*(1/2)*AB=(15/4)*AB.
From here we express the length of side AB: AB = (60*4)/15 = 16.
If the problem gives the lengths of two sides of a triangle and the angle between them, then you can apply the formula for the area of a triangle through the sine.
An example of calculating the area of a triangle using sine. Given sides are a = 3, b = 4, and angle γ = 30°. The sine of an angle of 30° is 0.5 
The area of the triangle will be 3 square meters. cm.
There may also be other conditions. If the length of one side and the angles are given, then first you need to calculate the missing angle. Because the sum of all angles of a triangle is 180°, then:
The area will be equal to half the square of the side multiplied by the fraction. Its numerator is the product of the sines of adjacent angles, and its denominator is the sine of the opposite angle. Now we calculate the area using the following formulas:
For example, given a triangle with side a=3 and angles γ=60°, β=60°. Calculate the third angle: 
Substituting the data into the formula
We find that the area of the triangle is 3.87 square meters. cm.
II. Area of a triangle through cosine
To find the area of a triangle, you need to know the lengths of all sides. Using the cosine theorem, you can find unknown sides, and only then use them.
According to the cosine theorem, the square of the unknown side of a triangle is equal to the sum of the squares of the remaining sides minus twice the product of these sides and the cosine of the angle located between them.
From the theorem we derive formulas for finding the length of the unknown side:
Knowing how to find the missing side, having two sides and the angle between them, you can easily calculate the area. The formula for the area of a triangle through the cosine helps to quickly and easily find solutions to various problems.
An example of calculating the formula for the area of a triangle using cosine
Given a triangle with known sides a = 3, b = 4, and angle γ = 45°. First, let's find the missing side With. Cosine 45°=0.7. To do this, we substitute the data into the equation derived from the cosine theorem.
Now using the formula, we find
side a triangle can be detected not only along the perimeter and area, but also along a given side and angles. For this, trigonometric functions are used - sinus and co sinus. Problems involving their use are found in school course geometry, as well as in the university course of analytical geometry and linear algebra.
Instructions
1. If you know one of the sides of a triangle and the angle between it and its other side, use trigonometric functions - sinus om and co sinus ohm Imagine a right triangle, NBC, whose angle is? equal to 60 degrees. The NBC triangle is shown in the figure. Because of sinus, as is well known, is the ratio of the opposite side to the hypotenuse, and sinus– the ratio of the adjacent leg to the hypotenuse; to solve the problem, use the following relationship between these parameters: sin ? = НB/BC Accordingly, if you want to find out the leg of a right triangle, express it through the hypotenuse in the following way: НB = BC*sin?
2. If in the problem statement, on the contrary, a leg of a triangle is given, find its hypotenuse, guided by the further relationship between the given quantities: BC = НB/sin? By analogy, find the sides of the triangle and using co sinus and, changing the previous expression as follows: cos ? = HC/BC
3. In elementary mathematics there is a representation of the theorem sinus ov. Guided by the facts that this theorem describes, it is also possible to discover the sides of the triangle. In addition, it allows you to detect the sides of a triangle inscribed in a circle if you know the radius of the latter. To do this, use the relationship given below:a/sin ?=b/sin b=c/sin y=2RThis theorem is applicable when two sides and an angle of a triangle are known, or one of the angles of a triangle and the radius of a circle circumscribed around it are given. .
4. Besides the theorem sinus ov, there is also a theorem that is essentially similar to it sinus ov, which, like the previous one, is also applicable to triangles of all 3 varieties: rectangular, acute and obtuse. Guided by the facts that prove this theorem, it is possible to find unknown quantities using the following relationships between them: c^2=a^2+b^2-2ab*cos?
Geometric figure consisting of three points, not belonging to one line, called vertices, and three pairwise segments connecting them, called sides, is called a triangle. There are a lot of problems for finding the sides and angles of a triangle given a limited number of initial data, one of these problems is finding the side of a triangle given one of its sides and two corners .
Instructions
1. Let a triangle be constructed?ABC and the famous side BC and angles?? and??. It is known that the sum of the angles of any triangle is 180?, therefore in the triangle? ABC is the angle?? will be equal?? = 180? – (?? + ??).The sides AC and AB can be found using the sine theorem, which says AB/sin?? = BC/sin?? = AC/sin?? = 2 * R, where R is the radius of the circle circumscribed about the triangle? ABC, then we get R = BC/sin??, AB = 2 * R * sin??, AC = 2 * R * sin??. The sine theorem can be used when all sorts of data on 2 corners and sides.
2. The sides of a given triangle can be found by calculating its area using the formula S = 2 * R? *sin?? *sin?? * sin??, where R is calculated by the formula R = BC/sin??, R is the radius of the circumscribed triangle? ABC from here Then side AB can be detected by calculating the height lowered on ith = BC * sin??, then using the formula S = 1/2 * h * AB we haveAB = 2 * S/hIn a similar way, it is possible to calculate side AC.
3. If the external angles of a triangle are given as angles?? and??, then it is possible to detect interior angles with the support of the corresponding relations?? = 180? – ??,?? = 180? – ??,?? = 180? – (?? + ??). Then we proceed similarly to the first two points.
The understanding of triangles has been carried out by mathematicians for several thousand years. The science of triangles - trigonometry - uses special quantities: sine and cosine.
Right triangle
Initially, sine and cosine arose from the need to calculate quantities in right triangles. It was noted that if the degree measure of angles in a right triangle is not changed, then the aspect ratio, no matter how much these sides change in length, remains invariably identical. This is how the representations of sine and cosine were introduced. The sine of an acute angle in a right triangle is the ratio of the opposite side to the hypotenuse, and the cosine is the ratio of the side adjacent to the hypotenuse.
Theorems of cosines and sines
But cosines and sines can be used for more than just right triangles. In order to discover the value of an obtuse or acute angle, the side of any triangle, it is enough to apply the theorem of cosines and sines. The cosine theorem is quite primitive: “The square of a side of a triangle is equal to the sum of the squares of the other 2 sides minus the double product of these sides by the cosine of the angle between them.” There are two interpretations of the sine theorem: small and extended. According to the minor: “In a triangle, the angles are proportional to the opposite sides.” This theorem is often expanded due to the property of the circumscribed circle of a triangle: “In a triangle, the angles are proportional to the opposite sides, and their ratio is equal to the diameter of the circumscribed circle.”
Derivatives
The derivative is a mathematical tool that shows how rapidly a function changes with respect to the metamorphosis of its argument. Derivatives are used in algebra, geometry, economics and physics, a number of technical disciplines. When solving problems, you need to know the tabular values of the derivatives of trigonometric functions: sine and cosine. The derivative of a sine is a cosine, and a cosine is a sine, but with a minus sign.
Application in mathematics
Sines and cosines are especially often used when solving right triangles and problems related to them. The convenience of sines and cosines is also reflected in technology. It was primitive to evaluate angles and sides using the theorems of cosines and sines, breaking down difficult shapes and objects into “primitive” triangles. Engineers and architects, often dealing with calculations of aspect ratios and degrees, spent a lot of time and effort calculating the cosines and sines of non-tabular angles. Then Bradis tables came to the rescue, containing thousands of values of sines, cosines, tangents and cotangents of various angles. IN Soviet time some teachers forced their students to memorize pages of Bradis tables.