If a line makess angle `90^(@), 60^(@)`, and theta with x, y and z-axis respectively, then acute `theta` = ............. .

Answer

To solve the problem, we need to find the acute angle \( \theta \) that a line makes with the z-axis, given that it makes angles of \( 90^\circ \) with the x-axis and \( 60^\circ \) with the y-axis.

### Step-by-Step Solution:

1. **Understanding Direction Cosines**:
- The angles that a line makes with the coordinate axes are related to direction cosines. If a line makes angles \( \alpha \), \( \beta \), and \( \gamma \) with the x, y, and z axes respectively, then the direction cosines \( L \), \( M \), and \( N \) are given by:
\[ L = \cos(\alpha), \quad M = \cos(\beta), \quad N = \cos(\gamma) \]

2. **Assigning Values**:
- From the problem, we know:
- \( \alpha = 90^\circ \) (angle with x-axis)
- \( \beta = 60^\circ \) (angle with y-axis)
- \( \gamma = \theta \) (angle with z-axis)

3. **Calculating Direction Cosines**:
- For \( \alpha = 90^\circ \):
\[ L = \cos(90^\circ) = 0 \]
- For \( \beta = 60^\circ \):
\[ M = \cos(60^\circ) = \frac{1}{2} \]
- For \( \gamma = \theta \):
\[ N = \cos(\theta) \]

4. **Using the Relation of Direction Cosines**:
- The relationship among the direction cosines is given by:
\[ L^2 + M^2 + N^2 = 1 \]
- Substituting the values we have:
\[ 0^2 + \left(\frac{1}{2}\right)^2 + N^2 = 1 \]
- This simplifies to:
\[ 0 + \frac{1}{4} + N^2 = 1 \]
- Therefore:
\[ N^2 = 1 - \frac{1}{4} = \frac{3}{4} \]
- Taking the square root gives:
\[ N = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} \]

5. **Determining \( \theta \)**:
- Since \( N = \cos(\theta) \), we have:
\[ \cos(\theta) = \frac{\sqrt{3}}{2} \]
- The angle \( \theta \) that corresponds to \( \cos(\theta) = \frac{\sqrt{3}}{2} \) is:
\[ \theta = 30^\circ \]

### Final Answer:
The acute angle \( \theta \) is \( 30^\circ \).

---