`alpha` is an acute angle.
(sin `alpha + cos alpha`) is

Options

A. greater than 1.
B. less than 1.
C. equal to 1.
D. (We cannot say any of these as it depends on the value of `alpha`)

Answer

To solve the problem of finding the value of \( \sin \alpha + \cos \alpha \) for an acute angle \( \alpha \), we will evaluate it step by step.

### Step 1: Understanding the Range of \( \alpha \)
Since \( \alpha \) is an acute angle, it lies in the range:
\[ 0 < \alpha < 90^\circ \]

### Step 2: Using the Trigonometric Values
We will calculate \( \sin \alpha + \cos \alpha \) for specific values of \( \alpha \) within the acute angle range. Let's take three common angles: \( 30^\circ \), \( 45^\circ \), and \( 60^\circ \).

### Step 3: Calculate for \( \alpha = 45^\circ \)
For \( \alpha = 45^\circ \):
\[ \sin 45^\circ = \frac{1}{\sqrt{2}}, \quad \cos 45^\circ = \frac{1}{\sqrt{2}} \]
Thus,
\[ \sin 45^\circ + \cos 45^\circ = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.414 \]

### Step 4: Calculate for \( \alpha = 30^\circ \)
For \( \alpha = 30^\circ \):
\[ \sin 30^\circ = \frac{1}{2}, \quad \cos 30^\circ = \frac{\sqrt{3}}{2} \]
Thus,
\[ \sin 30^\circ + \cos 30^\circ = \frac{1}{2} + \frac{\sqrt{3}}{2} = \frac{1 + \sqrt{3}}{2} \approx 1.366 \]

### Step 5: Calculate for \( \alpha = 60^\circ \)
For \( \alpha = 60^\circ \):
\[ \sin 60^\circ = \frac{\sqrt{3}}{2}, \quad \cos 60^\circ = \frac{1}{2} \]
Thus,
\[ \sin 60^\circ + \cos 60^\circ = \frac{\sqrt{3}}{2} + \frac{1}{2} = \frac{\sqrt{3} + 1}{2} \approx 1.366 \]

### Summary of Values
- For \( \alpha = 45^\circ \), \( \sin \alpha + \cos \alpha \approx 1.414 \)
- For \( \alpha = 30^\circ \), \( \sin \alpha + \cos \alpha \approx 1.366 \)
- For \( \alpha = 60^\circ \), \( \sin \alpha + \cos \alpha \approx 1.366 \)

### Conclusion
From the calculations, we can conclude that for acute angles, \( \sin \alpha + \cos \alpha \) is always greater than 1 and can reach a maximum value of \( \sqrt{2} \) when \( \alpha = 45^\circ \).