Math Problem Statement

Work out the size of angle CDA in the trapezium ABCD, giving your answer to 1 decimal place.

Solution

To solve the problem of finding the angle CDA\angle CDA in the trapezium ABCDABCD, let's break down the information provided and use trigonometry.

Given Information:

  • ABCDABCD is a trapezium with the following dimensions:
    • AB=6.5cmAB = 6.5 \, \text{cm} (vertical height from AA to BB)
    • BC=9cmBC = 9 \, \text{cm} (top side)
    • CD=6cmCD = 6 \, \text{cm} (diagonal connecting CC to DD)
    • AD=23cmAD = 23 \, \text{cm} (bottom side)

Steps to Find CDA\angle CDA:

  1. Draw a Perpendicular Line:

    • Drop a perpendicular from point CC to line ADAD, and label the point of intersection as EE.
    • This creates a right triangle CEDCED where CECE is the vertical height from CC to ADAD, and EDED is the horizontal distance between points DD and EE.
  2. Calculate CECE:

    • Since CECE is the same height as ABAB, we have CE=6.5cmCE = 6.5 \, \text{cm}.
  3. Calculate EDED:

    • EDED is found by subtracting BCBC from ADAD: ED=ADBC=23cm9cm=14cmED = AD - BC = 23 \, \text{cm} - 9 \, \text{cm} = 14 \, \text{cm}
  4. Use the Tangent Function to Find CDA\angle CDA:

    • In CED\triangle CED, tan(CDA)=CEED\tan(\angle CDA) = \frac{CE}{ED}: tan(CDA)=6.514\tan(\angle CDA) = \frac{6.5}{14} tan(CDA)0.4643\tan(\angle CDA) \approx 0.4643
  5. Calculate CDA\angle CDA:

    • Find the inverse tangent (arctan) of 0.4643: CDA=tan1(0.4643)25.0\angle CDA = \tan^{-1}(0.4643) \approx 25.0^\circ
    • Therefore, CDA25.0\angle CDA \approx 25.0^\circ (to 1 decimal place).

Final Answer:

25.0\boxed{25.0^\circ}

Would you like further details or explanations on any part of this solution? Here are some additional related questions:

  1. How can the cosine rule be applied to solve angle problems in trapeziums?
  2. What other trigonometric identities are useful for solving angles in non-right triangles?
  3. Can the sine rule be applied in this problem instead of tangent?
  4. How does the Pythagorean theorem assist in solving problems involving right triangles in trapeziums?
  5. What are other properties of trapeziums that might help in similar geometric problems?

Tip: Always label all points and sides when solving geometry problems, as it makes it easier to apply trigonometric functions correctly.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Geometry
Right-Angled Triangles

Formulas

Tangent function: tan(θ) = opposite / adjacent
Inverse Tangent function: θ = tan^(-1)(opposite / adjacent)

Theorems

Trigonometric Ratios in Right-Angled Triangles

Suitable Grade Level

Grade 10