In the xy-plane, a right triangle OAP is formed by the points \(\mathrm{O(0,0)}\), \(\mathrm{A(80,0)}\), and \(\mathrm{P(80,150)}\). Triangle OAP is s...

1. VISUALIZE the problem setup

• Given information:
  • Triangle OAP has vertices \(\mathrm{O(0,0), A(80,0), P(80,150)}\)
  • Triangle \(\mathrm{OAP \sim OBQ}\) with correspondence \(\mathrm{O \leftrightarrow O, A \leftrightarrow B, P \leftrightarrow Q}\)
  • Need to find \(\mathrm{cos(\angle OQB)}\)

• VISUALIZE the triangle: Drawing or imagining triangle OAP shows it's a right triangle with the right angle at A, since A is directly below P.

2. SIMPLIFY to find the triangle's side lengths

• The legs are:
  • \(\mathrm{OA = 80}\) (horizontal distance)
  • \(\mathrm{AP = 150}\) (vertical distance)

• SIMPLIFY using Pythagorean theorem to find hypotenuse:
  \(\mathrm{OP^2 = 80^2 + 150^2 = 6400 + 22500 = 28900}\)
  \(\mathrm{OP = \sqrt{28900} = 170}\)

Note: This is the \(\mathrm{8-15-17}\) Pythagorean triple scaled by 10, making the calculation cleaner

3. INFER the angle relationship from similarity

• Since triangle \(\mathrm{OAP \sim OBQ}\) with the given correspondence:
  • Angle at O corresponds to angle at O
  • Angle at A corresponds to angle at B
  • Angle at P corresponds to angle at Q

• INFER the key insight: \(\mathrm{\angle OPA}\) in triangle OAP corresponds to \(\mathrm{\angle OQB}\) in triangle OBQ
  Therefore: \(\mathrm{\angle OQB = \angle OPA}\)

4. APPLY trigonometry to find the cosine

• In triangle OAP, for angle OPA:
  • Adjacent side = \(\mathrm{AP = 150}\)
  • Hypotenuse = \(\mathrm{OP = 170}\)

• \(\mathrm{cos(\angle OPA) = \frac{adjacent}{hypotenuse} = \frac{150}{170} = \frac{15}{17}}\)

• Since \(\mathrm{\angle OQB = \angle OPA}\): \(\mathrm{cos(\angle OQB) = \frac{15}{17}}\)

Answer: C) \(\mathrm{\frac{15}{17}}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students fail to recognize that corresponding angles in similar triangles are equal, so they try to work directly with the unknown triangle OBQ instead of using the given triangle OAP.

Without this key insight, students may attempt to set up proportions or try to find coordinates for points B and Q, leading to unnecessary complexity and confusion. This often causes them to get stuck and guess randomly.

Second Most Common Error:

Incorrect trigonometry application: Students correctly identify that \(\mathrm{\angle OQB = \angle OPA}\) but then use the wrong sides when calculating the cosine ratio.

They might calculate \(\mathrm{cos(\angle OPA) = \frac{OA}{OP} = \frac{80}{170} = \frac{8}{17}}\), confusing the adjacent and opposite sides for the angle. This leads them to select Choice A (\(\mathrm{\frac{8}{17}}\)).

The Bottom Line:

This problem combines triangle similarity with trigonometry, requiring students to first recognize angle correspondence and then correctly identify which sides to use in the cosine calculation. The key breakthrough is realizing you can work entirely within the given triangle OAP.