Finding Angles A And B Solution To Trigonometric Equations

#Introduction

In trigonometry, we often encounter problems where we need to find the values of angles given certain trigonometric relationships. This article delves into such a problem, where we are given the values of the tangent of the sum and difference of two angles, A and B, and we are tasked with finding the individual values of A and B. Specifically, we are given that $\tan (A+B) = \sqrt{3}$ and $\tan (A-B) = \frac{1}{\sqrt{3}}$, with the constraints that $0^{\circ} < A+B \leq 90^{\circ}$ and $A > B$. This problem not only tests our understanding of trigonometric identities and values but also our ability to solve simultaneous equations. We will walk through the steps to solve this problem, providing explanations and insights along the way. By the end of this article, you will have a clear understanding of how to approach similar problems and a solid grasp of the underlying trigonometric principles. Solving this problem involves several key concepts, including the values of trigonometric functions for standard angles, the properties of the tangent function, and the solution of simultaneous equations. Understanding these concepts is crucial for success in trigonometry and related fields. So, let's dive in and explore how to find the values of angles A and B in this intriguing problem.

Understanding the Problem

Before we jump into the solution, let's make sure we fully understand the problem statement. We are given two equations involving the tangent function: $\tan (A+B) = \sqrt{3}$ and $\tan (A-B) = \frac{1}{\sqrt{3}}$. These equations relate the sum and difference of two angles, A and B, to specific trigonometric values. The constraints $0^{\circ} < A+B \leq 90^{\circ}$ and $A > B$ are also crucial. The first constraint tells us that the sum of the angles, A+B, lies in the first quadrant, where the tangent function is positive. The second constraint, $A > B$, tells us that angle A is greater than angle B, which will be important when we interpret our solutions. Now, the core of the problem is to find the individual values of A and B that satisfy both equations and the given constraints. This means we need to use our knowledge of trigonometric values and algebraic techniques to isolate A and B. The tangent function, in particular, has specific values for standard angles like 30°, 45°, and 60°, which we will use to our advantage. Recognizing these standard values and how they relate to the tangent function is a critical step in solving this problem. Moreover, the constraint $0^{\circ} < A+B \leq 90^{\circ}$ limits the possible solutions, as it confines A+B to the first quadrant. This constraint is essential because the tangent function has the same value for angles that differ by multiples of 180°, so without this constraint, we might have multiple sets of solutions. Therefore, by carefully considering the equations and constraints, we can develop a strategy to find the unique values of A and B that satisfy all conditions. Understanding the problem thoroughly is the first step towards a successful solution.

Solving for A + B

Now, let's focus on the first equation: $\tan (A+B) = \sqrt{3}$. To find the value of A+B, we need to recall the standard angles for which the tangent function equals $\sqrt{3}$. From our knowledge of trigonometric values, we know that $\tan 60^{\circ} = \sqrt{3}$. However, it's important to remember that the tangent function has a period of 180°, meaning that $\tan (60^{\circ} + 180^{\circ}n) = \sqrt{3}$ for any integer n. But, we have the constraint $0^{\circ} < A+B \leq 90^{\circ}$, which limits the possible values of A+B. Therefore, we only consider the angle in the first quadrant where the tangent is $\sqrt{3}$. This leads us to the conclusion that $A+B = 60^{\circ}$. This step is crucial because it simplifies the problem significantly. Instead of dealing with a trigonometric equation, we now have a simple algebraic equation relating A and B. This equation will be one of the two equations we need to solve for A and B. The other equation will come from the second given trigonometric equation. By recognizing the standard trigonometric value and applying the given constraint, we have successfully found the value of A+B. This demonstrates the importance of knowing standard trigonometric values and how to apply constraints to narrow down the possible solutions. The ability to quickly recall these values and constraints is a valuable skill in trigonometry and will help us solve the rest of the problem efficiently. Now that we have one equation, let's move on to the next one and see how we can find another equation relating A and B.

Solving for A - B

Next, let's consider the second equation: $\tan (A-B) = \frac{1}{\sqrt{3}}$. Similar to the previous step, we need to recall the standard angles for which the tangent function equals $\frac{1}{\sqrt{3}}$. We know that $\tan 30^{\circ} = \frac{1}{\sqrt{3}}$. Again, the tangent function has a period of 180°, but we need to consider the constraint $A > B$, which implies that $A-B$ must be a positive angle. Since the tangent function is positive in the first and third quadrants, we could potentially have solutions in both quadrants. However, we are looking for the simplest solution, and since we don't have an upper bound on $A-B$, we'll assume that $A-B$ is also in the first quadrant. Thus, we can conclude that $A-B = 30^{\circ}$. This gives us our second equation relating A and B. Now, we have two equations: $A+B = 60^{\circ}$ and $A-B = 30^{\circ}$. These are two linear equations in two variables, which we can solve using standard algebraic techniques. The process of finding the value of $A-B$ is similar to that of finding $A+B$. It requires recognizing the standard trigonometric value and applying any given constraints. In this case, the constraint $A > B$ is crucial, as it ensures that $A-B$ is a positive angle. Without this constraint, we might need to consider other possible solutions. By carefully considering the equation and the constraint, we have successfully found the value of $A-B$. This further simplifies our problem, as we now have two linear equations that we can solve simultaneously to find the values of A and B. The next step is to use these equations to find the individual values of A and B.

Solving the Simultaneous Equations

Now that we have the two equations, $A+B = 60^{\circ}$ and $A-B = 30^{\circ}$, we can solve them simultaneously to find the values of A and B. There are several methods to solve simultaneous equations, such as substitution, elimination, and matrix methods. In this case, the elimination method is particularly straightforward. We can add the two equations together to eliminate B:

$(A+B) + (A-B) = 60^{\circ} + 30^{\circ}$

This simplifies to:

$2A = 90^{\circ}$

Dividing both sides by 2, we get:

$A = 45^{\circ}$

Now that we have the value of A, we can substitute it into either of the original equations to find the value of B. Let's use the first equation, $A+B = 60^{\circ}$:

$45^{\circ} + B = 60^{\circ}$

Subtracting $45^{\circ}$ from both sides, we get:

$B = 15^{\circ}$

Therefore, we have found that $A = 45^{\circ}$ and $B = 15^{\circ}$. It's a good practice to check if these values satisfy both original equations and the given constraints. We can verify that $\tan (45^{\circ} + 15^{\circ}) = \tan 60^{\circ} = \sqrt{3}$ and $\tan (45^{\circ} - 15^{\circ}) = \tan 30^{\circ} = \frac{1}{\sqrt{3}}$. Also, $0^{\circ} < 45^{\circ} + 15^{\circ} = 60^{\circ} \leq 90^{\circ}$ and $45^{\circ} > 15^{\circ}$, so the constraints are satisfied. This confirms that our solution is correct. Solving simultaneous equations is a fundamental skill in mathematics, and this problem demonstrates how it can be applied in trigonometry. By using the elimination method, we were able to efficiently find the values of A and B. This step is the culmination of our efforts, bringing together the results from the previous steps to find the final solution.

Conclusion

In this article, we successfully found the values of angles A and B given the equations $\tan (A+B) = \sqrt{3}$ and $\tan (A-B) = \frac{1}{\sqrt{3}}$, with the constraints $0^{\circ} < A+B \leq 90^{\circ}$ and $A > B$. We found that $A = 45^{\circ}$ and $B = 15^{\circ}$. This problem demonstrates the importance of understanding trigonometric values for standard angles, the properties of trigonometric functions, and the ability to solve simultaneous equations. We started by understanding the problem and the given constraints. Then, we used the values of the tangent function for standard angles to find the values of $A+B$ and $A-B$. Finally, we solved the resulting simultaneous equations to find the individual values of A and B. The key steps in solving this problem were recognizing the standard trigonometric values, applying the given constraints, and using algebraic techniques to solve the equations. This problem is a good example of how different mathematical concepts can come together to solve a single problem. It also highlights the importance of a systematic approach to problem-solving. By breaking down the problem into smaller, more manageable steps, we were able to find the solution efficiently. This approach can be applied to a wide range of mathematical problems. In conclusion, understanding trigonometric functions and their properties, along with algebraic techniques, is crucial for solving problems like this. By mastering these concepts, you can confidently tackle similar problems and deepen your understanding of mathematics. The solution we found satisfies all the given conditions, confirming its correctness. This reinforces the importance of verifying solutions to ensure they are valid.