Finding Largest Angle In Triangle And Partial Fraction Decomposition
In this comprehensive guide, we will delve into two intriguing mathematical problems. First, we will explore a triangle with sides of specific lengths, aiming to determine the value of its largest angle. Then, we will tackle the decomposition of a rational expression into partial fractions. These problems, while seemingly distinct, showcase fundamental concepts in trigonometry and algebra, respectively. By understanding the underlying principles and applying appropriate techniques, we can effectively solve such challenges. This exploration will not only enhance our problem-solving skills but also deepen our appreciation for the elegance and interconnectedness of mathematical ideas.
Understanding the Problem
To find the largest angle of a triangle, when given the lengths of its sides, is a classic problem in trigonometry. The key to solving this lies in the Law of Cosines, a fundamental relationship between the sides and angles of any triangle. This law allows us to calculate the cosine of an angle if we know the lengths of all three sides. The largest angle in a triangle is always opposite the longest side. Therefore, our initial step is to identify the longest side among the given lengths. Once we know the longest side, we can apply the Law of Cosines to determine the angle opposite to it, which will be the largest angle of the triangle. This method provides a direct and efficient way to solve this type of problem, highlighting the power of trigonometric principles in geometric applications.
Identifying the Longest Side
In the given triangle, the sides have lengths (3 + √2) cm, 2√3 cm, and 3√2 cm. To determine the longest side, we need to compare these lengths. It's helpful to approximate the values of the square roots to make the comparison easier. We know that √2 is approximately 1.414, and √3 is approximately 1.732. Thus, the side lengths are approximately:
- 3 + √2 ≈ 3 + 1.414 = 4.414 cm
- 2√3 ≈ 2 * 1.732 = 3.464 cm
- 3√2 ≈ 3 * 1.414 = 4.242 cm
Comparing these approximations, we can see that (3 + √2) cm is the longest side. This is because 4.414 is greater than both 3.464 and 4.242. Therefore, the largest angle in the triangle will be the angle opposite the side with length (3 + √2) cm. Identifying the longest side is a crucial first step, as it directs us to the angle we need to calculate using the Law of Cosines. This process of approximation and comparison is a valuable skill in solving geometric problems.
Applying the Law of Cosines
The Law of Cosines is a powerful tool in trigonometry that relates the sides and angles of a triangle. It states that for any triangle with sides a, b, and c, and angles A, B, and C opposite those sides, the following equation holds:
c² = a² + b² - 2ab cos(C)
In our case, we want to find the angle opposite the longest side, (3 + √2) cm. Let's denote this angle as θ. The sides adjacent to this angle are 2√3 cm and 3√2 cm. Applying the Law of Cosines, we have:
(3 + √2)² = (2√3)² + (3√2)² - 2(2√3)(3√2) cos(θ)
Expanding and simplifying this equation will allow us to isolate cos(θ) and subsequently find the value of θ. This application of the Law of Cosines demonstrates its utility in solving for unknown angles in triangles, especially when side lengths are known. The formula provides a direct link between the geometry of the triangle and the trigonometric relationships of its angles.
Solving for the Angle
Now, let's solve the equation we derived from the Law of Cosines:
(3 + √2)² = (2√3)² + (3√2)² - 2(2√3)(3√2) cos(θ)
First, we expand each term:
(3 + √2)² = 9 + 6√2 + 2 = 11 + 6√2
(2√3)² = 4 * 3 = 12
(3√2)² = 9 * 2 = 18
So, the equation becomes:
11 + 6√2 = 12 + 18 - 12√6 cos(θ)
11 + 6√2 = 30 - 12√6 cos(θ)
Now, we isolate the term with cos(θ):
12√6 cos(θ) = 30 - 11 - 6√2
12√6 cos(θ) = 19 - 6√2
Next, we solve for cos(θ):
cos(θ) = (19 - 6√2) / (12√6)
To find the value of θ, we take the inverse cosine (arccos) of this value:
θ = arccos((19 - 6√2) / (12√6))
Approximating the value, we get:
θ ≈ arccos((19 - 6 * 1.414) / (12 * 2.449))
θ ≈ arccos((19 - 8.484) / 29.388)
θ ≈ arccos(10.516 / 29.388)
θ ≈ arccos(0.358)
θ ≈ 69°
Therefore, the largest angle of the triangle is approximately 69 degrees. This calculation demonstrates the step-by-step process of applying the Law of Cosines and using trigonometric functions to find unknown angles in a triangle. The final answer is a tangible result of the application of these mathematical principles.
Understanding Partial Fractions
Partial fraction decomposition is a technique used to express a rational function (a fraction where the numerator and denominator are polynomials) as a sum of simpler fractions. This is particularly useful in calculus for integrating rational functions, as the simpler fractions are often easier to integrate. The basic idea is to reverse the process of adding fractions with different denominators. We start with a single fraction and break it down into multiple fractions, each with a simpler denominator. This decomposition allows us to analyze and manipulate complex rational expressions more easily, making it a valuable tool in various mathematical contexts.
Setting up the Decomposition
We are given the rational expression (2x - 1) / (3x² + 4x + 1). The first step in partial fraction decomposition is to factor the denominator. The denominator 3x² + 4x + 1 can be factored as (3x + 1)(x + 1). This factorization is crucial because it determines the form of the partial fractions. Since we have two distinct linear factors in the denominator, we can express the original fraction as a sum of two fractions, each with one of these factors as its denominator. Thus, we set up the decomposition as follows:
(2x - 1) / (3x² + 4x + 1) = A / (3x + 1) + B / (x + 1)
where A and B are constants that we need to determine. This setup is the foundation for solving the problem, as it transforms the original expression into a form where we can solve for the unknown coefficients. The process of factoring the denominator and setting up the decomposition is a fundamental skill in partial fraction decomposition.
Solving for the Constants
To find the constants A and B, we need to eliminate the fractions in the equation:
(2x - 1) / ((3x + 1)(x + 1)) = A / (3x + 1) + B / (x + 1)
We can do this by multiplying both sides of the equation by the common denominator, which is (3x + 1)(x + 1):
2x - 1 = A(x + 1) + B(3x + 1)
Now, we have a polynomial equation. There are two common methods to solve for A and B: the substitution method and the method of equating coefficients.
Substitution Method:
We can choose specific values of x that will eliminate one of the unknowns. For example, if we let x = -1, the term with B will vanish:
2(-1) - 1 = A(-1 + 1) + B(3(-1) + 1)
-3 = 0 + B(-2)
B = 3/2
Similarly, if we let x = -1/3, the term with A will vanish:
2(-1/3) - 1 = A(-1/3 + 1) + B(3(-1/3) + 1)
-5/3 = A(2/3) + 0
A = -5/2
Equating Coefficients Method:
We can expand the equation and group the terms with the same powers of x:
2x - 1 = Ax + A + 3Bx + B
2x - 1 = (A + 3B)x + (A + B)
Now, we equate the coefficients of the corresponding terms:
Coefficient of x: A + 3B = 2
Constant term: A + B = -1
Solving this system of linear equations, we get A = -5/2 and B = 3/2.
Both methods lead to the same values for A and B. The substitution method is often quicker when the denominator has linear factors, while the method of equating coefficients is more generally applicable. Choosing the appropriate method can streamline the solution process. The values of A and B are the key to expressing the original rational function as a sum of partial fractions.
Final Decomposition
Now that we have found the values of A and B, we can write the final partial fraction decomposition:
(2x - 1) / (3x² + 4x + 1) = A / (3x + 1) + B / (x + 1)
Substituting A = -5/2 and B = 3/2, we get:
(2x - 1) / (3x² + 4x + 1) = (-5/2) / (3x + 1) + (3/2) / (x + 1)
To make the expression cleaner, we can rewrite the fractions:
(2x - 1) / (3x² + 4x + 1) = -5 / (2(3x + 1)) + 3 / (2(x + 1))
This is the partial fraction decomposition of the given rational expression. We have successfully expressed the original fraction as a sum of two simpler fractions. This decomposition is a valuable result, as it can be used to simplify further operations, such as integration or finding limits. The final decomposition demonstrates the power of this technique in breaking down complex expressions into manageable components.
In this comprehensive guide, we have explored two distinct yet interconnected mathematical problems. First, we successfully determined the largest angle of a triangle with given side lengths by applying the Law of Cosines. This showcased the importance of trigonometric principles in solving geometric problems. Then, we tackled the decomposition of a rational expression into partial fractions, a technique crucial in calculus and algebraic manipulations. By factoring the denominator, setting up the decomposition, and solving for the constants, we expressed the original fraction as a sum of simpler fractions. These examples highlight the breadth and depth of mathematical concepts and their applications in problem-solving. Understanding these techniques not only enhances our mathematical proficiency but also provides a foundation for tackling more complex challenges in various fields.
