Solving For Angles A And B With Trigonometric Equations And Constraints

In trigonometry, solving for unknown angles given certain conditions is a common and essential task. This article delves into a specific problem where we are given two equations involving trigonometric functions of angles A and B, along with constraints on their ranges and relationships. Our goal is to determine the values of angles A and B that satisfy these conditions. The problem presents a blend of algebraic manipulation and trigonometric identities, requiring a step-by-step approach to arrive at the solution. This exploration will not only demonstrate the solution process but also highlight the underlying principles of trigonometry and problem-solving strategies applicable to similar scenarios.

The problem at hand involves the equation $P^2(A + B) = \sqrt{3}$ and $\tan(A - B) = \frac{1}{\sqrt{3}}$, with the constraints $0^{\circ} < A + B \leq 90^{\circ}$ and $A > B \geq 0^{\circ}$. These conditions provide a defined space within which we can seek our solutions. By carefully analyzing the equations and the constraints, we can systematically narrow down the possibilities and pinpoint the unique values of A and B that meet all requirements. This journey through the problem will serve as an exercise in applying trigonometric knowledge to solve real-world problems.

We are given the following conditions:

1. $P^2(A + B) = \sqrt{3}$
2. $\tan(A - B) = \frac{1}{\sqrt{3}}$
3. $0^{\circ} < A + B \leq 90^{\circ}$
4. $A > B \geq 0^{\circ}$

Analyzing the Equations:

The first equation involves a trigonometric function which we assume is a typo and should be corrected to $\tan(A + B) = \sqrt{3}$. This assumption aligns with the context of trigonometric problems and allows for a meaningful solution. The second equation involves the tangent of the difference of the angles, $\tan(A - B)$, which is set to $\frac{1}{\sqrt{3}}$. These two equations, combined with the given constraints, form the basis for our solution.

The constraints provide essential boundaries for our solutions. The condition $0^{\circ} < A + B \leq 90^{\circ}$ limits the sum of the angles to the first quadrant, simplifying the possible solutions for trigonometric functions. The condition $A > B \geq 0^{\circ}$ ensures that both angles are non-negative and that A is strictly greater than B, preventing trivial or duplicate solutions. These constraints are crucial in narrowing down the solutions and ensuring that we arrive at a unique and valid answer.

Let's first address the equation $\tan(A + B) = \sqrt{3}$. We need to find the angle $(A + B)$ whose tangent is $\sqrt{3}$. Recall the values of the tangent function for standard angles:

• $\tan(0^{\circ}) = 0$
• $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$
• $\tan(45^{\circ}) = 1$
• \tan(60^{\circ}) = \sqrt{3}}
• $\tan(90^{\circ})$ is undefined

From this, we can see that $\tan(60^{\circ}) = \sqrt{3}$. Therefore, one possible solution for $A + B$ is $60^{\circ}$. Since the tangent function has a period of $180^{\circ}$, other solutions would be $60^{\circ} + 180^{\circ}n$, where n is an integer. However, given the constraint $0^{\circ} < A + B \leq 90^{\circ}$, the only valid solution is $A + B = 60^{\circ}$.

Determining the Solution within Constraints:

The constraint $0^{\circ} < A + B \leq 90^{\circ}$ is crucial here. It restricts the possible values of $A + B$ to the first quadrant. While there are infinitely many angles whose tangent is $\sqrt{3}$, only one of them falls within this range. This highlights the importance of constraints in solving trigonometric equations, as they help to narrow down the solutions and identify the specific values that meet all conditions. Without this constraint, we would have an infinite set of solutions for $A + B$, making it impossible to uniquely determine the values of A and B.

Next, let's consider the equation $\tan(A - B) = \frac{1}{\sqrt{3}}$. We need to find the angle $(A - B)$ whose tangent is $\frac{1}{\sqrt{3}}$. Again, we refer to the standard values of the tangent function:

• $\tan(0^{\circ}) = 0$
• $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$
• $\tan(45^{\circ}) = 1$
• \tan(60^{\circ}) = \sqrt{3}}
• $\tan(90^{\circ})$ is undefined

We find that $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$. Therefore, one solution for $A - B$ is $30^{\circ}$. Similar to the previous case, the tangent function has a period of $180^{\circ}$, so other solutions would be $30^{\circ} + 180^{\circ}n$, where n is an integer. However, we need to consider the constraint $A > B \geq 0^{\circ}$, which implies that $A - B$ must be positive and less than $90^{\circ}$. Thus, the only valid solution is $A - B = 30^{\circ}$.

Applying the Constraint A > B ≥ 0°:

The constraint $A > B \geq 0^{\circ}$ plays a significant role in determining the solution for $A - B$. It ensures that the difference between the angles is positive, which is crucial for the uniqueness of the solution. If we didn't have this constraint, we might consider negative values or values greater than $90^{\circ}$, which would lead to different solutions or no solutions at all. The condition $A > B$ directly implies that $A - B$ must be positive, while $B \geq 0^{\circ}$ ensures that both angles are non-negative. These constraints are essential for defining the solution space and arriving at the correct answer.

Now we have a system of two linear equations:

1. $A + B = 60^{\circ}$
2. $A - B = 30^{\circ}$

We can solve this system using various methods, such as substitution or elimination. Let's use the elimination method. Adding the two equations, we get:

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

$2A = 90^{\circ}$

$A = 45^{\circ}$

Now, substitute the value of A into the first equation:

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

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

$B = 15^{\circ}$

Therefore, the solution is $A = 45^{\circ}$ and $B = 15^{\circ}$.

Verification and Uniqueness of the Solution:

To ensure the correctness of our solution, we must verify that the values of A and B satisfy all the given conditions. First, let's check the original equations:

1. $\tan(A + B) = \tan(45^{\circ} + 15^{\circ}) = \tan(60^{\circ}) = \sqrt{3}$ (satisfied)
2. $\tan(A - B) = \tan(45^{\circ} - 15^{\circ}) = \tan(30^{\circ}) = \frac{1}{\sqrt{3}}$ (satisfied)

Next, we verify the constraints:

1. $0^{\circ} < A + B = 60^{\circ} \leq 90^{\circ}$ (satisfied)
2. $A = 45^{\circ} > B = 15^{\circ} \geq 0^{\circ}$ (satisfied)

Since all conditions are met, our solution is valid. Furthermore, the systematic approach we followed ensures that this is the unique solution to the problem. By carefully considering the constraints and solving the equations step-by-step, we have arrived at a definitive answer.

In conclusion, we have successfully found the values of angles A and B that satisfy the given trigonometric equations and constraints. By carefully analyzing the equations, applying trigonometric identities, and considering the constraints, we determined that $A = 45^{\circ}$ and $B = 15^{\circ}$. This problem demonstrates the importance of a systematic approach in solving trigonometric problems, as well as the crucial role of constraints in narrowing down the solutions.

This exercise highlights the interconnectedness of algebra and trigonometry. The problem required not only a strong understanding of trigonometric functions and their values but also the ability to solve a system of linear equations. The constraints acted as boundaries, guiding us towards the unique solution within the infinite possibilities. The solution process involved a blend of analytical thinking and mathematical manipulation, showcasing the power of combining different mathematical tools to solve complex problems.