Problem 15 - Algebra 2 and Trigonometry

Question

In \(15-20,\) find, to the nearest degree, the measure of an acute angle for which the given equation is true. $$ \sin \theta+3=5 \sin \theta $$

Step-by-Step Solution

Answer

\(\theta \approx 49°\)

Step 1: Simplifying the Equation

Start by moving all terms involving \(\sin \theta\) to one side of the equation: \(\sin \theta + 3 = 5 \sin \theta\). Subtract \(\sin \theta\) from both sides to get \(3 = 4\sin \theta\).

Step 2: Solving for Sine

Divide both sides of the equation by 4 to isolate \(\sin \theta\): \(\frac{3}{4} = \sin \theta\). This gives \(\sin \theta = 0.75\).

Step 3: Finding the Angle

To find \(\theta\), use the inverse sine function: \(\theta = \sin^{-1}(0.75)\).

Step 4: Rounding the Angle

Calculate \(\theta\) using a calculator: \(\theta \approx 48.5903°\). Round to the nearest degree: \(\theta \approx 49°\).

Key Concepts

Acute Angles

An acute angle is an angle that is less than 90 degrees. In trigonometry, we often deal with acute angles because they are part of right triangles. Understanding acute angles is essential when solving trigonometric equations, as they help determine the correct quadrant for angle measurements. In the problem we're looking at, the acute angle is involved in determining the value of \( \theta \) that satisfies a trigonometric equation. As per the original exercise, once we find \( \theta \) using the inverse sine functionality, it automatically gives us the acute angle measure since the inverse sine function only returns angles between \(-90^\circ\) and \(90^\circ\). This means our solution angle \( \theta \), which approximates to \(49^\circ\), comfortably qualifies as an acute angle.

Sine Function

The sine function is one of the most fundamental trigonometric functions. It's usually abbreviated as \( \sin \) and is essential for calculating relationships within right-angled triangles. For any given angle \( \theta \), the sine function returns a ratio. In a right triangle, it's the length of the opposite side over the hypotenuse.

Here is what sine tells us:

• It ranges between \(-1\) and \(1\).
• For acute angles (angles less than \(90^\circ\)), the sine values are positive.
• The sine function is periodic with a period of \(360^\circ\) (or \(2\pi\) radians).

In our problem, the equation \( \sin \theta = 0.75 \) tells us that the sine of angle \( \theta \) is \(0.75\). Because sine values for acute angles are positive, this reinforces that \( \theta \) is an acute angle.

Inverse Trigonometric Functions

Inverse trigonometric functions allow us to find angles when we know their trigonometric ratios. For the sine function, the inverse, denoted \( \sin^{-1} \) or arcsin, reveals the angle \( \theta \) whose sine is a given value. This function is extremely helpful when solving for angles in trigonometric equations.

Key features include:

• \( \sin^{-1} \) function returns angles within \(-90^\circ\) to \(90^\circ\) (or \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\) in radians).
• This means any angle we compute using \( \sin^{-1} \) will automatically give us an acute angle, provided the sine is positive.

In the given problem, after determining \( \sin \theta = 0.75 \), we used the inverse sine function to calculate the angle. By performing \( \theta = \sin^{-1}(0.75) \), we obtained approximately \(48.5903^\circ\), which reveals \( \theta \) as an acute angle once rounded to the nearest degree to \(49^\circ\).