In right triangle PQR, the right angle is at vertex Q. The lengths of the sides are such that the...

1. TRANSLATE the problem information

• Given information:
  • Right triangle PQR with right angle at Q
  • \(\sin(\mathrm{P}) = \frac{2\mathrm{x} + 7}{15}\)
  • \(\cos(\mathrm{R}) = \frac{5\mathrm{x} - 2}{15}\)
  • Angles P and R are acute

2. INFER the key relationship

• Since Q is the right angle (90°), angles P and R are the two remaining angles
• In any triangle, angles sum to 180°: \(\mathrm{P} + \mathrm{R} + 90° = 180°\)
• Therefore: \(\mathrm{P} + \mathrm{R} = 90°\), making them complementary angles
• Key insight: For complementary angles, \(\sin(\mathrm{P}) = \cos(\mathrm{R})\)

3. TRANSLATE this relationship into an equation

• Set the given expressions equal: \(\frac{2\mathrm{x} + 7}{15} = \frac{5\mathrm{x} - 2}{15}\)

4. SIMPLIFY to solve for x

• Since denominators are equal, numerators must be equal:
  \(2\mathrm{x} + 7 = 5\mathrm{x} - 2\)
• Collect like terms:
  \(7 + 2 = 5\mathrm{x} - 2\mathrm{x}\)
• Simplify:
  \(9 = 3\mathrm{x}\)
• Divide by 3:
  \(\mathrm{x} = 3\)

5. APPLY CONSTRAINTS to verify the solution

• Check that \(\mathrm{x} = 3\) gives valid trigonometric values:
  • \(\sin(\mathrm{P}) = \frac{2(3) + 7}{15} = \frac{13}{15} \approx 0.867\) ✓
  • \(\cos(\mathrm{R}) = \frac{5(3) - 2}{15} = \frac{13}{15} \approx 0.867\) ✓
• Both values are between 0 and 1, confirming validity

Answer: B (x = 3)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students fail to recognize that the acute angles in a right triangle are complementary, missing the crucial \(\sin(\mathrm{P}) = \cos(\mathrm{R})\) relationship.

Without this insight, they might try to use the Pythagorean theorem or other approaches that don't lead to a solvable equation with the given information. This leads to confusion and guessing among the answer choices.

Second Most Common Error:

Inadequate SIMPLIFY execution: Students correctly set up the equation \(\frac{2\mathrm{x} + 7}{15} = \frac{5\mathrm{x} - 2}{15}\) but make algebraic errors when solving.

A common mistake is: \(2\mathrm{x} + 7 = 5\mathrm{x} - 2 \rightarrow 7 - 2 = 5\mathrm{x} - 2\mathrm{x} \rightarrow 5 = 3\mathrm{x} \rightarrow \mathrm{x} = \frac{5}{3}\), which isn't an answer choice. This may lead them to select Choice A (1) as the closest option, or cause confusion and guessing.

The Bottom Line:

This problem requires recognizing the special relationship between complementary angles in right triangles. Students who miss this connection cannot progress systematically and resort to guessing.