Find the acute angle which the line with direction -cosines `lt (1)/(sqrt(3)), (1)/(sqrt(6)), n gt ` makes with positive direction of z-axis.

Answer

To find the acute angle that the line with direction cosines \(\left(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{6}}, n\right)\) makes with the positive direction of the z-axis, we can follow these steps:

### Step 1: Understand the Direction Cosines
The direction cosines of a line are defined as the cosines of the angles that the line makes with the coordinate axes. Here, we have:
- \(l = -\frac{1}{\sqrt{3}}\)
- \(m = \frac{1}{\sqrt{6}}\)
- \(n\) is unknown and represents the cosine of the angle with the z-axis.

### Step 2: Use the Property of Direction Cosines
The sum of the squares of the direction cosines is equal to 1:
\[ l^2 + m^2 + n^2 = 1 \]
Substituting the values of \(l\) and \(m\):
\[ \left(-\frac{1}{\sqrt{3}}\right)^2 + \left(\frac{1}{\sqrt{6}}\right)^2 + n^2 = 1 \]

### Step 3: Calculate \(l^2\) and \(m^2\)
Calculating \(l^2\):
\[ l^2 = \left(-\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3} \]
Calculating \(m^2\):
\[ m^2 = \left(\frac{1}{\sqrt{6}}\right)^2 = \frac{1}{6} \]

### Step 4: Substitute and Solve for \(n^2\)
Now substituting \(l^2\) and \(m^2\) into the equation:
\[ \frac{1}{3} + \frac{1}{6} + n^2 = 1 \]
To combine the fractions, find a common denominator (which is 6):
\[ \frac{2}{6} + \frac{1}{6} + n^2 = 1 \]
\[ \frac{3}{6} + n^2 = 1 \]
\[ n^2 = 1 - \frac{3}{6} = \frac{3}{6} = \frac{1}{2} \]

### Step 5: Find \(n\)
Taking the square root:
\[ n = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Step 6: Find the Angle with the Z-axis
The angle \(\gamma\) that the line makes with the positive z-axis is given by:
\[ \cos \gamma = n = \frac{1}{\sqrt{2}} \]
To find \(\gamma\):
\[ \gamma = \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) \]
This corresponds to:
\[ \gamma = 45^\circ \]

### Final Answer
The acute angle that the line makes with the positive direction of the z-axis is \(45^\circ\).