In ` Delta ABC , angle C = 90 ^(@) , AB = 20 and BC = 12 . D ` is a point in side AC such that CD = 9 Taking angle BDC = x, find
` sin angle ABC `

Answer

To solve the problem step by step, we will follow the given information and apply the Pythagorean theorem to find the required sine of angle ABC.

### Step-by-Step Solution:

1. **Identify the Triangle and Given Values:**
- We have triangle ABC with angle C = 90°.
- The lengths are given as:
- AB (hypotenuse) = 20
- BC (one leg) = 12

2. **Apply the Pythagorean Theorem:**
- According to the Pythagorean theorem, in a right triangle:
\[ AB^2 = AC^2 + BC^2 \]
- Substitute the known values:
\[ 20^2 = AC^2 + 12^2 \]
- This simplifies to:
\[ 400 = AC^2 + 144 \]

3. **Solve for AC:**
- Rearranging the equation gives:
\[ AC^2 = 400 - 144 \]
- Calculate the right side:
\[ AC^2 = 256 \]
- Taking the square root:
\[ AC = \sqrt{256} = 16 \]

4. **Find Sine of Angle ABC:**
- The sine of an angle in a right triangle is given by the ratio of the opposite side to the hypotenuse:
\[ \sin(\angle ABC) = \frac{AC}{AB} \]
- Substitute the values we found:
\[ \sin(\angle ABC) = \frac{16}{20} \]
- Simplifying this fraction:
\[ \sin(\angle ABC) = \frac{4}{5} \]

### Final Answer:
\[ \sin(\angle ABC) = \frac{4}{5} \]