If ` sin ^(-1) {(1)/( 2i) ( z - 3)} ` be the angle of a triangle and if z = x + iy then

Options

A. x = 1, y = 3
B. `x = 3, 0 lt y le 2`
C. ` x = , y = 2 `
D. ` x + y = 1 `

Answer

To solve the problem, we need to analyze the expression given and find the conditions on \( z \) such that the angle represented by \( \sin^{-1} \left( \frac{1}{2i} (z - 3) \right) \) is valid for a triangle.

### Step-by-Step Solution:

1. **Understanding the Function**:
The expression \( \sin^{-1}(x) \) is defined for \( x \) in the range \([-1, 1]\). Therefore, we need to ensure that:
\[ -1 \leq \frac{1}{2i} (z - 3) \leq 1 \]

2. **Substituting \( z \)**:
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers. Then, substituting this into the inequality gives:
\[ -1 \leq \frac{1}{2i} ((x + iy) - 3) \leq 1 \]
Simplifying this, we have:
\[ -1 \leq \frac{1}{2i} (x - 3 + iy) \leq 1 \]

3. **Multiplying by \( 2i \)**:
To eliminate the fraction, we multiply the entire inequality by \( 2i \). Note that multiplying by \( i \) rotates the complex number by \( 90^\circ \) in the complex plane:
\[ -2i \leq x - 3 + iy \leq 2i \]

4. **Separating Real and Imaginary Parts**:
From the inequality, we can separate the real and imaginary parts:
- The real part: \( x - 3 \) must be between the real parts of \(-2i\) and \(2i\), which are both 0. Therefore:
\[ x - 3 = 0 \implies x = 3 \]
- The imaginary part: \( y \) must be between the imaginary parts of \(-2i\) and \(2i\):
\[ -2 \leq y \leq 2 \]

5. **Final Result**:
Thus, we conclude that:
\[ x = 3 \quad \text{and} \quad y \in [0, 2] \]
Therefore, the values of \( z \) can be expressed as:
\[ z = 3 + iy \quad \text{where} \quad 0 \leq y \leq 2 \]

### Conclusion:
The values of \( z \) that satisfy the conditions of the problem are:
\[ z = 3 + iy \quad \text{for} \quad y \in [0, 2] \]