In right triangle PQR, the right angle is at vertex Q. The length of the hypotenuse PR is 30 units,...

1. TRANSLATE the problem information

• Given information:
  • Right triangle PQR with right angle at vertex Q
  • Hypotenuse \(\mathrm{PR = 30}\) units
  • Angle \(\mathrm{P = 28°}\)
  • Area can be written as \(k \cdot \sin(28°) \cdot \cos(28°)\)
• Need to find: the value of constant k

2. INFER the solution approach

• To find the area, we need the lengths of the two legs (PQ and QR)
• Since we know the hypotenuse and one acute angle, we can use trigonometry
• The area formula will help us identify what k must equal

3. APPLY trigonometric ratios to find the legs

• From angle P, the opposite leg is QR:
  \(\sin(28°) = \frac{\mathrm{QR}}{30}\), so \(\mathrm{QR} = 30 \cdot \sin(28°)\)
• From angle P, the adjacent leg is PQ:
  \(\cos(28°) = \frac{\mathrm{PQ}}{30}\), so \(\mathrm{PQ} = 30 \cdot \cos(28°)\)

4. APPLY the area formula

• Area = \(\frac{1}{2} \times \mathrm{base} \times \mathrm{height} = \frac{1}{2} \times \mathrm{PQ} \times \mathrm{QR}\)
• Area = \(\frac{1}{2} \times (30 \cdot \cos(28°)) \times (30 \cdot \sin(28°))\)

5. SIMPLIFY to match the given form

• \(\mathrm{Area} = \frac{1}{2} \times 30 \times 30 \times \cos(28°) \times \sin(28°)\)
• \(\mathrm{Area} = \frac{1}{2} \times 900 \times \sin(28°) \times \cos(28°)\)
• \(\mathrm{Area} = 450 \cdot \sin(28°) \cdot \cos(28°)\)

6. INFER the value of k

• Comparing our result with \(k \cdot \sin(28°) \cdot \cos(28°)\)
• We see that \(k = 450\)

Answer: D) 450

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students may try to use the area formula directly without first finding the individual leg lengths. They might attempt to use Area = \(\frac{1}{2} \times \mathrm{hypotenuse} \times \mathrm{something}\), not realizing that the hypotenuse isn't a base or height in the standard area formula. This leads to confusion and guessing.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly set up the trigonometry and area formula but make arithmetic errors when combining terms. For instance, they might calculate \(\frac{1}{2} \times 900\) incorrectly as 225 instead of 450, leading them to select Choice C (225).

The Bottom Line:

This problem requires connecting trigonometry with area formulas - students must recognize that finding the legs first is the pathway to expressing area in the required form. The key insight is that both sine and cosine appear naturally when you find both legs of the right triangle.