Step 1: Understanding the Concept:
A line in 3D space makes angles $\alpha, \beta, \gamma$ with the X, Y, and Z coordinate axes respectively. The cosines of these angles are called the direction cosines of the line. A fundamental geometric property is that the sum of the squares of the direction cosines is always equal to 1. We use this to find the missing angle.
Step 2: Key Formula or Approach:
Identity for direction cosines: $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.
Given $\alpha = 45^\circ$ and $\beta = 60^\circ$, solve for $\cos\gamma$ and then determine the obtuse angle $\gamma$.
Step 3: Detailed Explanation:
Let the angles made by the line with X, Y, and Z axes be $\alpha, \beta$, and $\gamma$ respectively.
We are given:
$\alpha = 45^\circ \implies \cos\alpha = \cos(45^\circ) = \frac{1}{\sqrt{2}}$
$\beta = 60^\circ \implies \cos\beta = \cos(60^\circ) = \frac{1}{2}$
We use the fundamental relationship:
\[ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \]
Substitute the known values:
\[ \left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2\gamma = 1 \]
\[ \frac{1}{2} + \frac{1}{4} + \cos^2\gamma = 1 \]
Find a common denominator to add the fractions:
\[ \frac{2}{4} + \frac{1}{4} + \cos^2\gamma = 1 \]
\[ \frac{3}{4} + \cos^2\gamma = 1 \]
Isolate $\cos^2\gamma$:
\[ \cos^2\gamma = 1 - \frac{3}{4} = \frac{1}{4} \]
Take the square root of both sides:
\[ \cos\gamma = \pm \sqrt{\frac{1}{4}} = \pm \frac{1}{2} \]
This gives two possible principal angles for $\gamma$:
If $\cos\gamma = \frac{1}{2}$, then $\gamma = 60^\circ$ (acute angle).
If $\cos\gamma = -\frac{1}{2}$, then $\gamma = 120^\circ$ (obtuse angle).
The problem explicitly asks for the obtuse angle $\theta$ made with the Z-axis.
Therefore, we must choose the negative cosine value.
$\theta = \gamma = 120^\circ$.
Step 4: Final Answer:
The obtuse angle is $120^\circ$.