Thank you for the clarification. Let's solve the problem with \( \sin(\angle P) = \frac{12}{13} \) instead of \( \frac{1}{2} \).

### Step 1: Interpret \( \sin(\angle P) = \frac{12}{13} \)
The sine of an angle in a right triangle is given by:
\[
\sin(\theta) = \frac{\text{opposite side}}{\text{hypotenuse}}
\]
So, for \( \angle P \), the opposite side \( PR \) is 12 units for every 13 units of the hypotenuse \( PQ \).

Given:
- \( PQ = 48 \) (the hypotenuse of triangle \( PQR \))

### Step 2: Calculate \( PR \)
Since \( \frac{PR}{PQ} = \frac{12}{13} \), we can find \( PR \) as:
\[
PR = \frac{12}{13} \times 48 = \frac{576}{13} \approx 44.31 \, \text{units}
\]

### Step 3: Find \( RQ \) using the Pythagorean theorem
Using the Pythagorean theorem in triangle \( PQR \):
\[
PQ^2 = PR^2 + RQ^2
\]
\[
48^2 = \left(\frac{576}{13}\right)^2 + RQ^2
\]
\[
2304 = \left(\frac{576}{13}\right)^2 + RQ^2
\]

Now let's compute \( RQ \):

\[
RQ^2 = 2304 - \left(\frac{576}{13}\right)^2
\]
\[
RQ^2 = 2304 - \frac{331776}{169}
\]
\[
RQ^2 = \frac{388416 - 331776}{169} = \frac{56640}{169}
\]
\[
RQ = \frac{\sqrt{56640}}{13} \approx \frac{238.04}{13} \approx 18.31 \, \text{units}
\]

### Step 4: Determine \( SQ \)
Since \( SQR \) is similar to \( PQR \) (because \( \angle R \) is common, and both triangles are right triangles), the ratios of corresponding sides will be the same.

To find \( SQ \) using the similarity of triangles:
\[
\frac{SQ}{RQ} = \frac{SR}{PR}
\]

Given that \( SR \) is some fraction of \( PR \), if \( SR = k \times PR \), then:
\[
\frac{SQ}{18.31} = \frac{k \times PR}{44.31}
\]
\[
SQ = k \times 18.31 \times \frac{44.31}{44.31} = k \times 18.31
\]

Since \( k \) is not given directly, we'd need more information or context to determine \( k \).

However, if \( k = 1 \) (i.e., \( S \) coincides with \( P \)), then:
\[
SQ = 18.31 \, \text{units}
\]

But if \( S \) is positioned differently or a specific value for \( k \) is implied (often \( k = \frac{12}{13} \) in similar problems), we'd use:
\[
SQ = \frac{12}{13} \times 18.31 = \frac{219.72}{13} \approx 16.9 \, \text{units}
\]

Thus, depending on the context, \( SQ \) could be interpreted as \( \approx 16.9 \) units or \( \approx 18.31 \) units.

Would you like further clarification on this, or any other assistance?

---

### Related Questions:

1. What is the role of the Pythagorean theorem in right triangle problems?
2. How does the sine ratio relate to the triangle sides?
3. How does similarity help in solving right triangle problems?
4. What is the impact of the value of \( k \) on \( SQ \)?
5. How would \( SQ \) change if \( PQ \) were different?
6. Can cosine or tangent ratios be used similarly in such problems?
7. What if we were given \( SR \) directly—how would that simplify the problem?
8. How do you approximate square roots in calculations like this?

### Tip:
Always double-check the problem's given ratios and measurements—small differences can lead to significant changes in the final solution.

Learn how to solve a trigonometric problem involving a right triangle where the sine of angle P is given. Detailed steps include using the sine ratio, Pythagorean theorem, and similarity of triangles. Suitable for Grades 10-12.

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