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In the triangle \( \triangle PQR \), we are given:

- \( \angle PQR = 90^\circ \) (right angle at \(Q\)),
- \( \angle PRQ = 38^\circ \),
- \( PQ = x + 0.8 \) cm,
- \( PR = 5.2 \) cm.

We can use trigonometry to solve for \( x \), specifically using the sine or cosine rule.

Since \( \angle PRQ = 38^\circ \), and the side opposite this angle is \( PQ \) (which equals \( x + 0.8 \)), we can use the sine function to relate these values:

\[
\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}
\]

In this case:

\[
\sin(38^\circ) = \frac{PQ}{PR} = \frac{x + 0.8}{5.2}
\]

Let's calculate the value of \( x \).The value of \( x \) is approximately \( 2.40 \) cm.

Would you like a detailed explanation or have any other questions?

Here are some related questions:
1. How can you verify the value of \( x \) using a different trigonometric function?
2. What is the length of side \( QR \) in this triangle?
3. How do you calculate the area of the triangle \( PQR \)?
4. What is the cosine of \( \angle PRQ \), and how can it be used here?
5. Can you solve for \( x \) if the angle changes from \( 38^\circ \) to another value?

**Tip:** In right-angle triangles, the sine, cosine, and tangent ratios are particularly useful for finding unknown side lengths.

Find the value of x in the given right-angled triangle where angle PRQ = 38°, PR = 5.2 cm, and PQ = x + 0.8 cm.

Learn how to find the value of x using the sine rule in a right-angled triangle, where angle PRQ = 38°, PR = 5.2 cm, and PQ = x + 0.8 cm. This solution covers essential trigonometric concepts and the sine function for Grades 9-11.

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