Question:Triangle ABC is a right triangle with right angle at C. The altitude from C meets hypotenuse AB at point...

1. TRANSLATE the problem information

• Given information:
  • Right triangle ABC with right angle at C
  • Altitude CD from C meets hypotenuse AB at point D
  • \(\tan(\mathrm{A}) = \frac{50}{7}\)
• Find: \(\tan(\angle\mathrm{CBD})\)

2. INFER the complementary angle relationship

• In any right triangle, the two acute angles are complementary (sum to 90°)
• This means angles A and B satisfy: \(\mathrm{A} + \mathrm{B} = 90°\)
• For complementary angles, there's a key tangent relationship: \(\tan(\mathrm{A}) \times \tan(\mathrm{B}) = 1\)
• Therefore: \(\tan(\mathrm{B}) = \frac{1}{\tan(\mathrm{A})}\)

3. SIMPLIFY to find tan(B)

• Given \(\tan(\mathrm{A}) = \frac{50}{7}\)
• \(\tan(\mathrm{B}) = \frac{1}{\tan(\mathrm{A})} = \frac{1}{\frac{50}{7}} = \frac{7}{50}\)

4. INFER the triangle similarity relationship

• When you draw an altitude from the right angle to the hypotenuse, it creates two smaller triangles
• Both smaller triangles are similar to the original triangle and to each other
• Triangle CBD is similar to triangle ABC
• In this similarity, angle CBD corresponds to angle B of the original triangle

5. INFER the final relationship

• Since triangle CBD ~ triangle ABC, corresponding angles are equal
• Therefore: \(\angle\mathrm{CBD} = \angle\mathrm{B}\)
• This means: \(\tan(\angle\mathrm{CBD}) = \tan(\mathrm{B}) = \frac{7}{50}\)

Answer: 7/50

Why Students Usually Falter on This Problem

Most Common Error Path:

Missing conceptual knowledge: Students don't remember that \(\tan(\mathrm{A}) \times \tan(\mathrm{B}) = 1\) for complementary angles.

Without this relationship, they might try to use the given \(\tan(\mathrm{A}) = \frac{50}{7}\) directly or attempt complex coordinate geometry approaches. They get stuck trying to find angle measures or side lengths, leading to confusion and guessing.

Second Most Common Error:

Weak INFER skill: Students don't recognize that the altitude creates similar triangles with corresponding equal angles.

They might correctly find \(\tan(\mathrm{B}) = \frac{7}{50}\) but then think \(\angle\mathrm{CBD}\) is different from \(\angle\mathrm{B}\). This leads them to attempt trigonometric calculations within triangle CBD without recognizing the similarity, causing them to abandon the systematic solution and guess.

The Bottom Line:

This problem requires connecting two key geometric insights: the complementary angle relationship in right triangles and the similarity created by the altitude to the hypotenuse. Students who miss either connection cannot reach the elegant solution.