ABCD is a rectangle where ratio of the length of AB and BC is 3 : 2. If P is the mid point of AB . Find sin `angle CPB`

Answer

To find \( \sin \angle CPB \) in rectangle ABCD where the ratio of the lengths of sides AB and BC is 3:2, we can follow these steps:

### Step 1: Define the lengths of the rectangle
Let the length of \( AB \) be \( 3x \) and the length of \( BC \) be \( 2x \). Thus, we have:
- \( AB = 3x \)
- \( BC = 2x \)

### Step 2: Identify the coordinates of the points
Assuming point \( A \) is at the origin (0, 0):
- \( A(0, 0) \)
- \( B(3x, 0) \)
- \( C(3x, 2x) \)
- \( D(0, 2x) \)

### Step 3: Find the midpoint \( P \) of \( AB \)
The midpoint \( P \) of segment \( AB \) can be calculated as:
\[ P = \left( \frac{0 + 3x}{2}, \frac{0 + 0}{2} \right) = \left( \frac{3x}{2}, 0 \right) \]

### Step 4: Calculate the lengths of sides \( BC \) and \( PC \)
- The length \( BC \) is simply \( 2x \).
- To find the length \( PC \), we use the distance formula:
\[ PC = \sqrt{(3x - \frac{3x}{2})^2 + (2x - 0)^2} \]
Calculating this:
\[ PC = \sqrt{\left(\frac{3x}{2}\right)^2 + (2x)^2} = \sqrt{\frac{9x^2}{4} + 4x^2} = \sqrt{\frac{9x^2}{4} + \frac{16x^2}{4}} = \sqrt{\frac{25x^2}{4}} = \frac{5x}{2} \]

### Step 5: Apply the sine definition in triangle \( BCP \)
In triangle \( BCP \), we know:
\[ \sin \angle CPB = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{PC} \]
Substituting the values we found:
\[ \sin \angle CPB = \frac{2x}{\frac{5x}{2}} = \frac{2x \cdot 2}{5x} = \frac{4}{5} \]

### Conclusion
Thus, the value of \( \sin \angle CPB \) is:
\[ \sin \angle CPB = \frac{4}{5} \]

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