Find leg by angle and leg online. We find the side of a triangle if the other two are known in three ways, formulas
In life, we will often have to deal with mathematical problems: at school, at university, and then helping our child with 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 look at one of them: finding the side of a right triangle.
What is a right triangle
First, let's remember what a right triangle is. A right triangle is a 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 the right angle is called the 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: “The square of the hypotenuse is equal to the sum of the squares of the 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 a leg using trigonometric functions: 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.
The first are the segments that are adjacent to the right angle, and the hypotenuse is the longest part of the figure and is located opposite the angle of 90 degrees. A Pythagorean triangle is one whose sides are equal to the natural numbers; their lengths in this case are called “Pythagorean triple”.
Egyptian triangle
In order for the current generation to recognize geometry in the form in which it is taught in school now, it has developed over several centuries. The fundamental point is considered to be the Pythagorean theorem. The sides of a rectangular is known throughout the world) are 3, 4, 5.
Few people are not familiar with the phrase “Pythagorean pants are equal in all directions.” However, in reality the theorem sounds like this: c 2 (square of the hypotenuse) = a 2 + b 2 (sum of squares of the legs).
Among mathematicians, a triangle with sides 3, 4, 5 (cm, m, etc.) is called “Egyptian”. The interesting thing is that which is inscribed in the figure is equal to one. The name arose around the 5th century BC, when Greek philosophers traveled to Egypt.
When building the pyramids, architects and surveyors used the ratio 3:4:5. Such structures turned out to be proportional, pleasant to look at and spacious, and also rarely collapsed.
In order to build a right angle, the builders used a rope with 12 knots tied on it. In this case, the probability of constructing a right triangle increased to 95%.
Signs of equality of figures
- An acute angle in a right triangle and a long side, which are equal to the same elements in the second triangle, are an indisputable sign of equality of figures. Taking into account the sum of the angles, it is easy to prove that the second acute angles are also equal. Thus, the triangles are identical according to the second criterion.
- When superimposing two figures on top of each other, we rotate them so that, when combined, they become one isosceles triangle. According to its property, the sides, or more precisely, the hypotenuses, are equal, as well as the angles at the base, which means that these figures are the same.
Based on the first sign, it is very easy to prove that the triangles are indeed equal, the main thing is that the two smaller sides (i.e., the legs) are equal to each other.
The triangles will be identical according to the second criterion, the essence of which is the equality of the leg and the acute angle.
Properties of a triangle with a right angle
The height that is lowered from the right angle splits the figure into two equal parts.
The sides of a right triangle and its median can be easily recognized by the rule: the median that falls on the hypotenuse is equal to half of it. can be found both by Heron's formula and by the statement that it is equal to half the product of the legs.
In a right triangle, the properties of angles of 30°, 45° and 60° apply.
- With an angle of 30°, it should be remembered that the opposite leg will be equal to 1/2 of the largest side.
- If the angle is 45°, then the second acute angle is also 45°. This suggests that the triangle is isosceles and its legs are the same.
- The property of an angle of 60° is that the third angle has a degree measure of 30°.
The area can be easily found out using one of three formulas:
- through the height and the side on which it descends;
- according to Heron's formula;
- on the sides and the angle between them.
The sides of a right triangle, or rather the legs, converge with two altitudes. In order to find the third, it is necessary to consider the resulting triangle, and then, using the Pythagorean theorem, calculate the required length. In addition to this formula, there is also a relationship between twice the area and the length of the hypotenuse. The most common expression among students is the first one, as it requires fewer calculations.
Theorems applying to right triangle
Right triangle geometry involves the use of theorems such as:
The lengths of the sides (a, b, c) are known, use the cosine theorem. It states that the square of the length of any of the sides is equal to the sum of the squares of the lengths of the other two, from which twice the product of the lengths of the same two sides by the cosine of the angle between them is subtracted. You can use this theorem to calculate the angle at any of the vertices; it is important to know only its location relative to the sides. For example, to find the angle α that lies between sides b and c, the theorem must be written as follows: a² = b² + c² - 2*b*c*cos(α).
Express the cosine of the desired angle from the formula: cos(α) = (b²+c²-a²)/(2*b*c). To both sides of the equality, apply the inverse function of cosine - arc cosine. It allows you to restore the angle in degrees using the cosine value: arccos(cos(α)) = arccos((b²+c²-a²)/(2*b*c)). The left side can be simplified and the calculation of the angle between sides b and c will take the final form: α = arccos((b²+c²-a²)/2*b*c).
When finding the values of acute angles in a right triangle, knowing the lengths of all sides is not necessary; two of them are sufficient. If these two sides are legs (a and b), divide the length of the one opposite the desired angle (α) by the length of the other. This way you will get the tangent value of the desired angle tg(α) = a/b, and by applying the inverse function - arctangent - to both sides of the equality and simplifying the left side, as in the previous step, derive the final formula: α = arctan(a/b ).
If the known sides are the leg (a) and the hypotenuse (c), to calculate the angle (β) formed by these sides, use the cosine function and its inverse - arc cosine. The cosine is determined by the ratio of the length of the leg to the hypotenuse, and the formula in its final form can be written as follows: β = arccos(a/c). To calculate from the same initial acute angle (α) lying opposite the known leg, use the same relationship, replacing arccosine with arcsine: α = arcsin(a/c).
Sources:
- triangle formula with 2 sides
Tip 2: How to find the angles of a triangle by the lengths of its sides
There are several options for finding the values of all angles in a triangle if the lengths of its three are known parties. One way is to use two different formulas for calculating area triangle. To simplify calculations, you can also apply the sine theorem and the sum of angles theorem triangle.
Instructions
Use, for example, two formulas for calculating area triangle, one of which involves only three of his known parties s (Heron), and in the other - two parties s and the sine of the angle between them. Using different pairs in the second formula parties, you can determine the magnitude of each of the angles triangle.
Solve the problem in general form. Heron's formula determines the area triangle, as the square root of the product of the semi-perimeter (half of all parties) on the difference between the semi-perimeter and each of parties. If we replace with the sum parties, then the formula can be written as follows: S=0.25∗√(a+b+c)∗(b+c-a)∗(a+c-b)∗(a+b-c).C other parties s area triangle can be expressed as half the product of its two parties by the sine of the angle between them. For example, for parties a and b with an angle γ between them, this formula can be written as follows: S=a∗b∗sin(γ). Replace the left side of the equality with Heron's formula: 0.25∗√(a+b+c)∗(b+c-a)∗(a+c-b)∗(a+b-c)=a∗b∗sin(γ). Derive from this equality the formula for
Online calculator.
Solving triangles.
Solving a triangle is finding all its six elements (i.e., three sides and three angles) from any three given elements that define the triangle.
This mathematical program finds sides \(b, c\), and angle \(\alpha \) from user-specified side \(a\) and two adjacent angles \(\beta \) and \(\gamma \)
The program not only gives the answer to the problem, but also displays the process of finding a solution.
This online calculator can be useful for high school students in secondary schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.
In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.
If you are not familiar with the rules for entering numbers, we recommend that you familiarize yourself with them.
Rules for entering numbers
Numbers can be specified not only as whole numbers, but also as fractions.
The integer and fractional parts in decimal fractions can be separated by either a period or a comma.
For example, you can enter decimal fractions like 2.5 or like 2.5
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A little theory.
Theorem of sines
Theorem
The sides of a triangle are proportional to the sines of the opposite angles:
$$ \frac(a)(\sin A) = \frac(b)(\sin B) = \frac(c)(\sin C) $$
Cosine theorem
Theorem
Let AB = c, BC = a, CA = b in triangle ABC. Then
The square of a side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of those sides multiplied by the cosine of the angle between them.
$$ a^2 = b^2+c^2-2ba \cos A $$
Solving triangles
Solving a triangle means finding all its six elements (i.e., three sides and three angles) from any three given elements that define the triangle.
Let's look at three problems involving solving a triangle. In this case, we will use the following notation for the sides of triangle ABC: AB = c, BC = a, CA = b.
Solving a triangle using two sides and the angle between them
Given: \(a, b, \angle C\). Find \(c, \angle A, \angle B\)
Solution
1. Using the cosine theorem we find \(c\):
$$ \cos A = \frac( b^2+c^2-a^2 )(2bc) $$
3. \(\angle B = 180^\circ -\angle A -\angle C\)
Solving a triangle by side and adjacent angles
Given: \(a, \angle B, \angle C\). Find \(\angle A, b, c\)
Solution
1. \(\angle A = 180^\circ -\angle B -\angle C\)
$$ b = a \frac(\sin B)(\sin A), \quad c = a \frac(\sin C)(\sin A) $$
Solving a triangle using three sides
Given: \(a, b, c\). Find \(\angle A, \angle B, \angle C\)
Solution
1. Using the cosine theorem we obtain:
$$ \cos A = \frac(b^2+c^2-a^2)(2bc) $$
2. Similarly, we find angle B.
3. \(\angle C = 180^\circ -\angle A -\angle B\)
Solving a triangle using two sides and an angle opposite a known side
Given: \(a, b, \angle A\). Find \(c, \angle B, \angle C\)
Solution
1. Using the theorem of sines, we find \(\sin B\) we get:
$$ \frac(a)(\sin A) = \frac(b)(\sin B) \Rightarrow \sin B = \frac(b)(a) \cdot \sin A $$
Let's introduce the notation: \(D = \frac(b)(a) \cdot \sin A \). Depending on the number D, the following cases are possible:
If D > 1, such a triangle does not exist, because \(\sin B\) cannot be greater than 1
If D = 1, there is a unique \(\angle B: \quad \sin B = 1 \Rightarrow \angle B = 90^\circ \)
If D If D 2. \(\angle C = 180^\circ -\angle A -\angle B\)
3. Using the sine theorem, we calculate the side c:
$$ c = a \frac(\sin C)(\sin A) $$
