A Right Triangle Has Two Acute Angles, $\theta$ And $\phi$, Such That $\cos \phi = X + 0.2$ And $\sin \theta = 2x - 0.5$. What Is The Value Of $x$?A. 0.7 B. 0.07 C. 1.0 D. 0.1

Introduction

In trigonometry, right triangles are fundamental concepts that help us understand various mathematical relationships. Given a right triangle with two acute angles, $\theta$ and $\phi$, we can use trigonometric functions to relate the angles to the side lengths of the triangle. In this article, we will explore how to find the value of $x$ in a right triangle with $\cos \phi = x + 0.2$ and $\sin \theta = 2x - 0.5$.

Understanding the Trigonometric Functions

Before we dive into the problem, let's recall the definitions of the trigonometric functions sine and cosine. The sine of an angle $\theta$ is defined as the ratio of the length of the side opposite the angle to the length of the hypotenuse. The cosine of an angle $\phi$ is defined as the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

Using the Pythagorean Identity

We can use the Pythagorean identity to relate the sine and cosine of an angle. The Pythagorean identity states that $\sin^2 \theta + \cos^2 \theta = 1$. We can use this identity to find a relationship between the sine and cosine of the angles $\theta$ and $\phi$.

Finding the Relationship Between the Angles

Since we are given the values of $\cos \phi$ and $\sin \theta$, we can use the Pythagorean identity to find a relationship between the angles. We can start by squaring both sides of the equation $\cos \phi = x + 0.2$ to get $\cos^2 \phi = (x + 0.2)^2$. Similarly, we can square both sides of the equation $\sin \theta = 2x - 0.5$ to get $\sin^2 \theta = (2x - 0.5)^2$.

Using the Pythagorean Identity to Find the Relationship

Now, we can use the Pythagorean identity to find a relationship between the angles. We can substitute the expressions for $\cos^2 \phi$ and $\sin^2 \theta$ into the Pythagorean identity to get $(x + 0.2)^2 + (2x - 0.5)^2 = 1$.

Expanding and Simplifying the Equation

We can expand and simplify the equation $(x + 0.2)^2 + (2x - 0.5)^2 = 1$ to get $x^2 + 0.4x + 0.04 + 4x^2 - 2x + 0.25 = 1$. Combining like terms, we get $5x^2 + 0.3x + 0.29 = 1$.

Rearranging the Equation

We can rearrange the equation $5x^2 + 0.3x + 0.29 = 1$ to get $5x^2 + 0.3x - 0.71 = 0$.

Solving the Quadratic Equation

We can solve the quadratic equation $5x^2 + 0.3x - 0.71 = 0$ using the quadratic formula. The quadratic formula states that the solutions to the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Applying the Quadratic Formula

We can apply the quadratic formula to the equation $5x^2 + 0.3x - 0.71 = 0$ to get $x = \frac{-0.3 \pm \sqrt{(0.3)^2 - 4(5)(-0.71)}}{2(5)}$.

Simplifying the Expression

We can simplify the expression $x = \frac{-0.3 \pm \sqrt{(0.3)^2 - 4(5)(-0.71)}}{2(5)}$ to get $x = \frac{-0.3 \pm \sqrt{0.09 + 14.2}}{10}$.

Evaluating the Expression

We can evaluate the expression $x = \frac{-0.3 \pm \sqrt{0.09 + 14.2}}{10}$ to get $x = \frac{-0.3 \pm \sqrt{14.29}}{10}$.

Simplifying the Square Root

We can simplify the square root $\sqrt{14.29}$ to get $3.8$.

Evaluating the Expression

We can evaluate the expression $x = \frac{-0.3 \pm 3.8}{10}$ to get two possible values for $x$: $x = \frac{-0.3 + 3.8}{10}$ and $x = \frac{-0.3 - 3.8}{10}$.

Evaluating the First Expression

We can evaluate the first expression $x = \frac{-0.3 + 3.8}{10}$ to get $x = \frac{3.5}{10}$.

Simplifying the Fraction

We can simplify the fraction $\frac{3.5}{10}$ to get $x = 0.35$.

Evaluating the Second Expression

We can evaluate the second expression $x = \frac{-0.3 - 3.8}{10}$ to get $x = \frac{-4.1}{10}$.

Simplifying the Fraction

We can simplify the fraction $\frac{-4.1}{10}$ to get $x = -0.41$.

Checking the Solutions

We can check the solutions $x = 0.35$ and $x = -0.41$ to see if they satisfy the original equations. We can substitute $x = 0.35$ into the equations $\cos \phi = x + 0.2$ and $\sin \theta = 2x - 0.5$ to get $\cos \phi = 0.35 + 0.2 = 0.55$ and $\sin \theta = 2(0.35) - 0.5 = 0.3$. We can check if these values satisfy the Pythagorean identity.

Checking the Pythagorean Identity

We can check if the values $\cos \phi = 0.55$ and $\sin \theta = 0.3$ satisfy the Pythagorean identity. We can substitute these values into the Pythagorean identity to get $\cos^2 \phi + \sin^2 \theta = 0.55^2 + 0.3^2 = 0.3025 + 0.09 = 0.3925$. Since this value is not equal to 1, we can conclude that $x = 0.35$ is not a valid solution.

Checking the Second Solution

We can check the second solution $x = -0.41$ to see if it satisfies the original equations. We can substitute $x = -0.41$ into the equations $\cos \phi = x + 0.2$ and $\sin \theta = 2x - 0.5$ to get $\cos \phi = -0.41 + 0.2 = -0.21$ and $\sin \theta = 2(-0.41) - 0.5 = -1.32$. We can check if these values satisfy the Pythagorean identity.

Checking the Pythagorean Identity

We can check if the values $\cos \phi = -0.21$ and $\sin \theta = -1.32$ satisfy the Pythagorean identity. We can substitute these values into the Pythagorean identity to get $\cos^2 \phi + \sin^2 \theta = (-0.21)^2 + (-1.32)^2 = 0.0441 + 1.744 = 1.7881$. Since this value is not equal to 1, we can conclude that $x = -0.41$ is not a valid solution.

Conclusion

We have found that neither of the solutions $x = 0.35$ and $x = -0.41$ satisfy the original equations. However, we can try to find a solution by using a different approach.

Alternative Approach

We can try to find a solution by using a different approach. We can start by squaring both sides of the equation $\cos \phi = x + 0.2$ to get $\cos^2 \phi = (x + 0.2)^2$. Similarly, we can square both sides of the equation $\sin \theta = 2x - 0.5$ to get $\sin^2 \theta = (2x - 0.5)^2$.

Using the Pythagorean Identity

We can use the Pythagorean identity to find a relationship between the angles. We can substitute the expressions for $\cos^2 \phi$ and $\sin^2 \theta$ into the Pythagorean identity to get $(x + 0.2)^2 + (2x - 0.5)^2 = 1$.

Expanding and Simplifying the Equation

We can expand and simplify the equation $(x + 0.2)^2 + (2x - 0.5)^2 = 1$ to get $x^2 + 0.4x + 0.04 + 4x^2 - 2x + 0.25 = 1$. Combining

Introduction

In our previous article, we explored how to find the value of $x$ in a right triangle with $\cos \phi = x + 0.2$ and $\sin \theta = 2x - 0.5$. We used the Pythagorean identity to find a relationship between the angles and then solved the resulting quadratic equation to find the value of $x$. However, we found that neither of the solutions we obtained satisfied the original equations. In this article, we will provide a Q&A section to help clarify any questions or doubts that readers may have.

Q: What is the Pythagorean identity?

A: The Pythagorean identity is a fundamental concept in trigonometry that states that the sum of the squares of the sine and cosine of an angle is equal to 1. Mathematically, it can be expressed as $\sin^2 \theta + \cos^2 \theta = 1$.

Q: How do we use the Pythagorean identity to find a relationship between the angles?

A: We can use the Pythagorean identity to find a relationship between the angles by substituting the expressions for $\cos^2 \phi$ and $\sin^2 \theta$ into the Pythagorean identity. This will give us an equation that relates the two angles.

Q: What is the quadratic formula?

A: The quadratic formula is a mathematical formula that is used to solve quadratic equations. It states that the solutions to the quadratic equation $ax^2 + bx + c = 0$ are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Q: How do we use the quadratic formula to solve the quadratic equation?

A: We can use the quadratic formula to solve the quadratic equation by substituting the values of $a$, $b$, and $c$ into the formula. This will give us two possible solutions for the value of $x$.

Q: Why did we not obtain a valid solution for the value of $x$?

A: We did not obtain a valid solution for the value of $x$ because the solutions we obtained did not satisfy the original equations. This may be due to the fact that the quadratic equation we obtained was not solvable or that the solutions we obtained were not valid.

Q: What is the alternative approach to finding the value of $x$?

A: The alternative approach to finding the value of $x$ is to use a different method to solve the quadratic equation. This may involve using a different formula or technique to solve the equation.

Q: Can you provide an example of how to use the alternative approach?

A: Yes, we can provide an example of how to use the alternative approach. Let's say we want to find the value of $x$ in the equation $\cos \phi = x + 0.2$ and $\sin \theta = 2x - 0.5$. We can start by squaring both sides of the equation $\cos \phi = x + 0.2$ to get $\cos^2 \phi = (x + 0.2)^2$. Similarly, we can square both sides of the equation $\sin \theta = 2x - 0.5$ to get $\sin^2 \theta = (2x - 0.5)^2$. We can then use the Pythagorean identity to find a relationship between the angles and solve the resulting quadratic equation.

Q: What are some common mistakes to avoid when solving quadratic equations?

A: Some common mistakes to avoid when solving quadratic equations include:

• Not checking if the solutions satisfy the original equation
• Not using the correct formula or technique to solve the equation
• Not simplifying the equation before solving it
• Not checking if the solutions are valid

Q: How can we check if the solutions are valid?

A: We can check if the solutions are valid by substituting the values of $x$ into the original equation and checking if they satisfy the equation.

Q: What are some common applications of quadratic equations?

A: Some common applications of quadratic equations include:

• Finding the maximum or minimum value of a quadratic function
• Solving systems of linear equations
• Modeling real-world phenomena, such as the motion of an object under the influence of gravity

Conclusion

In this article, we provided a Q&A section to help clarify any questions or doubts that readers may have about finding the value of $x$ in a right triangle with $\cos \phi = x + 0.2$ and $\sin \theta = 2x - 0.5$. We also discussed some common mistakes to avoid when solving quadratic equations and provided some common applications of quadratic equations.