Problem 30 - Algebra 2 and Trigonometry

Question

A water balloon leaves the air cannon at an angle of \(\theta\) with the ground and an initial velocity of 40 feet per second. The water balloon lands 30 feet from the cannon. The distance \(d\) traveled by the water balloon is given by the formula $$ d=\frac{1}{32} v^{2} \sin 2 \theta $$ where \(v\) is the initial velocity. a. Let \(x=2 \theta .\) Solve the equation \(30=\frac{1}{32}(40)^{2} \sin x\) to the nearest tenth of a degree. b. Use the formula \(x=2 \theta\) and your answer to part a to find the measure of the angle that the cannon makes with the ground.

Step-by-Step Solution

Answer

The angle \( \theta \) is approximately 18.5 degrees.

Step 1: Simplify the given equation

We start with the equation provided: \( 30 = \frac{1}{32} (40)^2 \sin x \). Begin by simplifying the term \( \frac{1}{32} (40)^2 \). Calculate this part: \( \frac{1}{32} \times 1600 = 50 \). So, the equation becomes \( 30 = 50 \sin x \).

Step 2: Solve for \( \sin x \)

With the equation \( 30 = 50 \sin x \), divide both sides by 50 to isolate \( \sin x \): \( \sin x = \frac{30}{50} = 0.6 \).

Step 3: Find \( x \) using inverse sine

To find \( x \), take the inverse sine (arc sine) of 0.6. Using a calculator, \( x = \sin^{-1} (0.6) \approx 36.87 \) degrees.

Step 4: Round to the nearest tenth

Now, round \( x = 36.87 \) degrees to the nearest tenth. This gives us \( x \approx 36.9 \) degrees.

Step 5: Find \( \theta \) from \( x = 2\theta \)

Given that \( x = 2\theta \), we solve for \( \theta \) using \( \theta = \frac{x}{2} \). Substitute the value of \( x \) we found: \( \theta = \frac{36.9}{2} = 18.45 \) degrees.

Step 6: Round \( \theta \) to the nearest tenth

Finally, round \( \theta = 18.45 \) to the nearest tenth. So, \( \theta \approx 18.5 \) degrees.

Key Concepts

Inverse Functions

Inverse functions are crucial when you need to reverse the effect of another function. In trigonometry, the inverse sine function, often written as \( \sin^{-1} \) or "arc sine," is used to find an angle whose sine value is known. This helps us solve equations involving trigonometric ratios.

• Example: If you know that \( \sin x = 0.6 \) and you need to find \( x \), you use \( x = \sin^{-1}(0.6) \).
• Calculators or trigonometric tables can help you find these values.
• The result, \( x \approx 36.87 \) degrees, is rounded appropriately based on the context of your problem, as seen here with a rounding to the nearest tenth giving \( x \approx 36.9 \) degrees.

Angle of Projection

The angle of projection determines how an object like a water balloon travels. It is the angle at which an object is launched relative to the ground. This angle plays a crucial role in determining the object's trajectory and the distance it will cover.

• In our example, the angle is connected to the equation through \( x = 2\theta \).
• You first solve for \( x \) and then use the relationship \( \theta = \frac{x}{2} \) to find the angle of projection relative to the ground.
• This results in an easier analysis of projectile motion, such as determining distance traveled, using equations derived from physics and trigonometry.

Sine Function

The sine function is a fundamental part of trigonometry, dealing with the ratio of the length of the side opposite to an angle to the length of the hypotenuse in a right triangle. It's also represented in the unit circle and is periodic, meaning it repeats its values in regular intervals.

• In a unit circle, \( \sin \theta \) represents the y-coordinate of a point as the angle \( \theta \) changes.
• For projectile problems like ours, understanding \( \sin 2\theta \) is crucial as it's part of the formula used for calculating the distance traveled.
• \( \sin x \) values range between -1 and 1, and solving equations often involves finding angles that fulfill this range constraint.

Trigonometric Equation Solving

Solving trigonometric equations involves finding the angles that satisfy a given equation. These equations frequently involve sine, cosine, or tangent functions. The process usually requires:

• Identifying the trigonometric function within the equation.
• Isolating the trigonometric function as was done here: \( 30 = 50 \sin x \Rightarrow \sin x = 0.6 \).
• Using inverse functions to solve for the angle, ensuring the solution is within the specified range, often around the unit circle's typical 0 to 360 degrees or 0 to \(2\pi\) radians boundary.
• Rounding numerical results to the precision required, in this case, to the nearest tenth of a degree.