`alpha` and `beta` are the angle made by a vector from positive x & positive y-axes redpectively. Which set of `alpha` and `beta` is not possible

Options

A. `45^(@), 60^(@)`
B. `30^(@), 60^(@)`
C. `60^(@), 60^(@)`
D. `30^(@), 45^(@)`

Explanation

To solve the problem, we need to determine which set of angles \( \alpha \) and \( \beta \) does not satisfy the condition for a vector in a two-dimensional plane. The condition we will use is:

\[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 \]

where \( \gamma \) is the angle made with the z-axis. Since we are only concerned with \( \alpha \) and \( \beta \), we can rearrange this equation to find \( \cos^2 \gamma \):

\[ \cos^2 \gamma = 1 - \cos^2 \alpha - \cos^2 \beta \]

The value of \( \cos^2 \gamma \) must be between 0 and 1 (inclusive). Therefore, we need to check each set of angles provided in the options to see if they satisfy this condition.

### Step-by-step Solution:

1. **Option 1: \( \alpha = 45^\circ, \beta = 60^\circ \)**
- Calculate \( \cos^2 \alpha \) and \( \cos^2 \beta \):
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \Rightarrow \cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
\[ \cos 60^\circ = \frac{1}{2} \Rightarrow \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
- Substitute into the equation:
\[ \cos^2 \gamma = 1 - \left(\frac{1}{2} + \frac{1}{4}\right) = 1 - \frac{3}{4} = \frac{1}{4} \]
- Since \( \frac{1}{4} \) is valid, this option is possible.

2. **Option 2: \( \alpha = 30^\circ, \beta = 60^\circ \)**
- Calculate \( \cos^2 \alpha \) and \( \cos^2 \beta \):
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \Rightarrow \cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
\[ \cos 60^\circ = \frac{1}{2} \Rightarrow \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
- Substitute into the equation:
\[ \cos^2 \gamma = 1 - \left(\frac{3}{4} + \frac{1}{4}\right) = 1 - 1 = 0 \]
- Since \( 0 \) is valid, this option is possible.

3. **Option 3: \( \alpha = 60^\circ, \beta = 60^\circ \)**
- Calculate \( \cos^2 \alpha \) and \( \cos^2 \beta \):
\[ \cos 60^\circ = \frac{1}{2} \Rightarrow \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
\[ \cos 60^\circ = \frac{1}{2} \Rightarrow \cos^2 60^\circ = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]
- Substitute into the equation:
\[ \cos^2 \gamma = 1 - \left(\frac{1}{4} + \frac{1}{4}\right) = 1 - \frac{1}{2} = \frac{1}{2} \]
- Since \( \frac{1}{2} \) is valid, this option is possible.

4. **Option 4: \( \alpha = 30^\circ, \beta = 45^\circ \)**
- Calculate \( \cos^2 \alpha \) and \( \cos^2 \beta \):
\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \Rightarrow \cos^2 30^\circ = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
\[ \cos 45^\circ = \frac{1}{\sqrt{2}} \Rightarrow \cos^2 45^\circ = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]
- Substitute into the equation:
\[ \cos^2 \gamma = 1 - \left(\frac{3}{4} + \frac{1}{2}\right) = 1 - \left(\frac{3}{4} + \frac{2}{4}\right) = 1 - \frac{5}{4} = -\frac{1}{4} \]
- Since \( -\frac{1}{4} \) is not valid (it is less than 0), this option is not possible.

### Conclusion:
The set of angles \( \alpha = 30^\circ \) and \( \beta = 45^\circ \) is not possible.