Question

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Explanation:

Step1: Solve for \( x \) using \( \sin(38^\circ) \)

We start with the equation \( \sin(38^\circ) = \frac{x}{15} \). To isolate \( x \), we multiply both sides by 15:
\( 15 \cdot \sin(38^\circ) = \frac{x}{15} \cdot 15 \)
Simplifying the right side, \( \frac{x}{15} \cdot 15 = x \), so \( 15\sin(38^\circ) = x \).
Using a calculator (in degrees), \( \sin(38^\circ) \approx 0.6157 \), so \( 15 \cdot 0.6157 \approx 9.2355 \approx 9.2 \). Thus, \( 9.2 = x \).

Step2: Identify the angle for the cosine equation

In the right triangle \( \triangle ABC \), angle at \( A \) is \( 52^\circ \) (since the angles in a triangle sum to \( 180^\circ \), and \( 90^\circ + 38^\circ + 52^\circ = 180^\circ \)). For angle \( A = 52^\circ \), the adjacent side to \( 52^\circ \) is \( x \), and the hypotenuse is 15. The cosine of an angle in a right triangle is \( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \), so \( \cos(52^\circ) = \frac{x}{15} \). Thus, the angle in the cosine equation is \( 52^\circ \).

Step3: Solve for \( x \) using \( \cos(52^\circ) \)

Starting with \( \cos(52^\circ) = \frac{x}{15} \), multiply both sides by 15:
\( 15 \cdot \cos(52^\circ) = \frac{x}{15} \cdot 15 \)
Simplifying, \( 15\cos(52^\circ) = x \). Using a calculator, \( \cos(52^\circ) \approx 0.6157 \) (wait, no—wait, \( \cos(52^\circ) \approx 0.6157 \)? Wait, no, \( \sin(38^\circ) \approx \cos(52^\circ) \) because \( 38^\circ + 52^\circ = 90^\circ \), so \( \sin(\theta) = \cos(90^\circ - \theta) \). So \( \cos(52^\circ) = \sin(38^\circ) \approx 0.6157 \), so \( 15 \cdot 0.6157 \approx 9.2 \)? Wait, no, wait—wait, actually, let's recalculate \( \cos(52^\circ) \). Wait, \( 52^\circ \): \( \cos(52^\circ) \approx 0.6157 \), so \( 15 \cdot 0.6157 \approx 9.2355 \approx 9.2 \)? Wait, no, maybe I made a mistake. Wait, no—wait, the first equation was \( \sin(38^\circ) = \frac{x}{15} \), giving \( x \approx 9.2 \). The second equation is \( \cos(52^\circ) = \frac{x}{15} \), so solving for \( x \), \( x = 15\cos(52^\circ) \). Let's calculate \( \cos(52^\circ) \): using a calculator, \( 52^\circ \), cosine is approximately \( 0.6157 \), so \( 15 \times 0.6157 \approx 9.2355 \approx 9.2 \). Wait, but actually, \( \sin(38^\circ) \approx 0.6157 \), \( \cos(52^\circ) \approx 0.6157 \), so both give \( x \approx 9.2 \). Wait, but let's check with a calculator: \( \cos(52^\circ) \approx 0.6156614753 \), so \( 15 \times 0.6156614753 \approx 9.2349 \approx 9.2 \).

Filling in the blanks:

1. First blank (in \( (15)\sin(38^\circ) = \underline{\quad}(15) \)): The operation is multiplying both sides by 15, so the blank is \( \frac{x}{15} \cdot \) (wait, no—the original equation is \( \sin(38^\circ) = \frac{x}{15} \), so to solve for \( x \), we multiply both sides by 15. So the left side is \( 15\sin(38^\circ) \), and the right side is \( \frac{x}{15} \times 15 \). So the blank in \( (15)\sin(38^\circ) = \underline{\quad}(15) \) is \( \frac{x}{15} \times \), but actually, the structure is \( (15)\sin(38^\circ) = \underline{\quad}(15) \), so the blank is \( \frac{x}{15} \times \)? Wait, no—let's look at the equation: \( \sin(38^\circ) = \frac{x}{15} \). Multiply both sides by 15: \( 15 \cdot \sin(38^\circ) = \frac{x}{15} \cdot 15 \). So the left side is \( (15)\sin(38^\circ) \), and the right side is \( \frac{x}{15} \times 15 \). So the blank in \( (15)\sin(38^\circ) = \underline{\quad}(15) \) is \( \frac{x}{15} \times \), but the problem has \( (15)\sin(38^\circ) = \underline{\quad}(15) \), so the blank is \( \frac{x}{15} \times \), but the way it's written is \( (15)\sin(38^\circ) = \underline{\quad}(15) \), so the blank is \( \frac{x}{15} \times \), but maybe it's \( \frac{x}{15} \cdot \), so the blank is \( \frac{x}{15} \). Wait, no—the original step is \( (15)\sin(38^\circ) = \underline{\quad}(15) \), and then \( (15)\sin(38^\circ) = x \). So the right side is \( \frac{x}{15} \times 15 \), so the blank is \( \frac{x}{15} \times \), but the problem's format is \( (15)\sin(38^\circ) = \underline{\quad}(15) \), so the blank is \( \frac{x}{15} \). Then the next blank is \( x \), since \( 9.2 = x \). Then the cosine angle is \( 52 \), and the last blank is \( 15\cos(52^\circ) \approx 9.2 \) (wait, no—wait, \( \cos(52^\circ) \approx 0.6157 \), so \( 15 \times 0.6157 \approx 9.2 \), so the last blank is \( 9.2 \).

Wait, let's re-express the blanks:

1. First blank: In \( (15)\sin(38^\circ) = \underline{\quad}(15) \), the right side is \( \frac{x}{15} \times 15 \), so the blank is \( \frac{x}{15} \times \), but the problem has \( (15)\sin(38^\circ) = \underline{\quad}(15) \), so the blank is \( \frac{x}{15} \).

2. Second blank: \( 9.2 = \underline{\quad} \), so the blank is \( x \).

3. Third blank: \( \cos(\underline{\quad}^\circ) = \frac{x}{15} \), so the blank is \( 52 \).

4. Fourth blank: \( \underline{\quad} \approx x \), where \( x = 15\cos(52^\circ) \approx 9.2 \), so the blank is \( 9.2 \).

But maybe the first blank is the operation, but let's proceed with the trigonometric solution.

Answer:

• First blank (in \( (15)\sin(38^\circ) = \underline{\quad}(15) \)): \( \frac{x}{15} \)
• Second blank (in \( 9.2 = \underline{\quad} \)): \( x \)
• Third blank (in \( \cos(\underline{\quad}^\circ) = \frac{x}{15} \)): \( 52 \)
• Fourth blank (in \( \underline{\quad} \approx x \)): \( 9.2 \)

(Note: The first blank's context is multiplying both sides by 15, so the right side is \( \frac{x}{15} \times 15 \), hence the blank is \( \frac{x}{15} \).)