Solving Trigonometric Equations And Rectangle Dimensions

Solving $5 \cos \theta - 1 = 0$ where $0^{\circ} \leq \theta \leq 90^{\circ}$

Let's dive into solving this trigonometric equation step by step. Trigonometric equations might seem daunting at first, but with a clear approach, they become quite manageable. Our main goal here is to isolate the cosine function and then find the angle $ heta$ that satisfies the equation within the given range.

First, we have the equation $5 \cos \theta - 1 = 0$. To isolate the cosine term, we need to get it by itself on one side of the equation. We can start by adding 1 to both sides. This gives us $5 \cos \theta = 1$. Now, we need to get rid of the 5 that's multiplying the cosine. To do this, we divide both sides by 5, which gives us $\cos \theta = \frac{1}{5}$. Great! We've now isolated the cosine function.

Now that we have $\cos \theta = \frac{1}{5}$, we need to find the angle $\theta$ whose cosine is $\frac{1}{5}$. To do this, we use the inverse cosine function, often denoted as $\cos^{-1}$ or arccos. So, we have $\theta = \cos^{-1}(\frac{1}{5})$. You'll need a calculator for this step, guys! Make sure your calculator is in degree mode since we're looking for $\theta$ in degrees.

Using a calculator, we find that $\theta \approx 78.46^{\circ}$. Now, we need to check if this angle falls within our given range, which is $0^{\circ} \leq \theta \leq 90^{\circ}$. Since $78.46^{\circ}$ is indeed between 0 and 90 degrees, it is a valid solution. So, the solution to the equation $5 \cos \theta - 1 = 0$ in the given range is approximately $78.46^{\circ}$. Remember, solving trigonometric equations often involves isolating the trigonometric function and then using inverse trigonometric functions to find the angle. Always double-check that your solutions fall within the specified range.

Solving $\cos(\alpha + 14)^{\circ} = \sin(4\alpha + 6)^{\circ}$

This equation involves both cosine and sine functions, which adds a little twist. To tackle this, we'll use the cofunction identity which states that $\cos(x) = \sin(90^{\circ} - x)$. This identity is super useful because it allows us to express cosine in terms of sine, or vice versa. By using this identity, we can rewrite the equation so that both sides involve the same trigonometric function. This makes it much easier to solve.

Using the cofunction identity, we can rewrite the left side of the equation. We have $\cos(\alpha + 14)^{\circ}$, which can be rewritten as $\sin(90^{\circ} - (\alpha + 14)^{\circ})$. Simplifying this, we get $\sin(90^{\circ} - \alpha - 14^{\circ}) = \sin(76^{\circ} - \alpha)$. So, our equation now becomes $\sin(76^{\circ} - \alpha) = \sin(4\alpha + 6)^{\circ}$. Now that both sides of the equation have the sine function, we can proceed by equating the arguments (the expressions inside the sine functions).

Setting the arguments equal to each other, we have $76^{\circ} - \alpha = 4\alpha + 6^{\circ}$. This is a simple linear equation in terms of $\alpha$, which we can solve. Let's start by adding $\alpha$ to both sides to get $76^{\circ} = 5\alpha + 6^{\circ}$. Next, subtract $6^{\circ}$ from both sides to get $70^{\circ} = 5\alpha$. Finally, divide both sides by 5 to isolate $\alpha$. This gives us $\alpha = 14^{\circ}$.

However, guys, there's a catch! Sine functions have a periodic nature, meaning they repeat their values at regular intervals. The sine function has a period of $360^{\circ}$, and also $\sin(x) = \sin(180^{\circ} - x)$. So, we need to consider another possibility where the arguments are supplementary. This means we need to consider the case where $76^{\circ} - \alpha = 180^{\circ} - (4\alpha + 6)^{\circ}$.

Let's solve this second case. We have $76^\circ} - \alpha = 180^{\circ} - 4\alpha - 6^{\circ}$, which simplifies to $76^{\circ} - \alpha = 174^{\circ} - 4\alpha$. Adding $4\alpha$ to both sides gives $76^{\circ} + 3\alpha = 174^{\circ}$. Subtracting $76^{\circ}$ from both sides gives $3\alpha = 98^{\circ}$. Finally, dividing by 3 gives $\alpha = \frac{98}{3}^{\circ} \approx 32.67^{\circ}$. So, we have two possible solutions for $\alpha$ $14^{\circ$ and approximately $32.67^{\circ}$. When solving equations involving trigonometric functions, it's crucial to consider all possible solutions within the domain, taking into account the periodic nature and relevant identities.

Rectangle Floor Dimensions

Now, let's switch gears and tackle a geometry problem involving a rectangle floor. We're given that the sides of the rectangle are $x$ cm and $C$ cm. However, the problem statement ends abruptly here. To provide a comprehensive solution, we need more information or a specific question related to these dimensions. For example, we might need to find the area or perimeter of the rectangle, or we might have some constraints on the values of $x$ and $C$. Without a clear objective, we can only offer general formulas and relationships.

The area of a rectangle is given by the formula $A = length \times width$. In this case, the area would be $A = x \times C$ square centimeters. The perimeter of a rectangle is the total length of all its sides, which can be calculated using the formula $P = 2(length + width)$. For our rectangle, the perimeter would be $P = 2(x + C)$ centimeters.

If we had additional information, such as the value of the area or perimeter, or a relationship between $x$ and $C$, we could solve for specific values. For instance, if we were told that the area is 100 square centimeters, we would have the equation $x \times C = 100$. If we also knew the value of $x$, we could easily find $C$, or vice versa. Similarly, if we knew the perimeter, we could use the perimeter formula to establish another equation and potentially solve for the dimensions.

In summary, while we can express the area and perimeter of the rectangle in terms of $x$ and $C$, we need more information to find specific values for these dimensions. Remember, geometric problems often require applying formulas and relationships, and having sufficient information is key to finding a solution.