Let `x=(sum_(n=1)^(44) cos n^(@))/(sum_(n=1)^(44) sin n^(@))` , find the greatest integer that does not exceed.

Answer

To solve the problem, we need to find the value of

\[ x = \frac{\sum_{n=1}^{44} \cos n^\circ}{\sum_{n=1}^{44} \sin n^\circ} \]

### Step 1: Calculate the sum of cosines
The sum of cosines can be expressed using the formula for the sum of cosines of an arithmetic series:

\[ \sum_{n=1}^{N} \cos n^\circ = \frac{\sin\left(\frac{N \cdot d}{2}\right) \cdot \cos\left(a + \frac{(N-1) \cdot d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]

where \(N\) is the number of terms, \(d\) is the common difference (which is \(1^\circ\) in this case), and \(a\) is the first term (which is \(1^\circ\)).

For \(N = 44\), \(d = 1^\circ\), and \(a = 1^\circ\):

\[ \sum_{n=1}^{44} \cos n^\circ = \frac{\sin\left(\frac{44 \cdot 1^\circ}{2}\right) \cdot \cos\left(1^\circ + \frac{(44-1) \cdot 1^\circ}{2}\right)}{\sin\left(\frac{1^\circ}{2}\right)} \]

Calculating the values:

\[ = \frac{\sin(22^\circ) \cdot \cos(22.5^\circ)}{\sin(0.5^\circ)} \]

### Step 2: Calculate the sum of sines
Similarly, for the sum of sines:

\[ \sum_{n=1}^{N} \sin n^\circ = \frac{\sin\left(\frac{N \cdot d}{2}\right) \cdot \sin\left(a + \frac{(N-1) \cdot d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]

For \(N = 44\), \(d = 1^\circ\), and \(a = 1^\circ\):

\[ \sum_{n=1}^{44} \sin n^\circ = \frac{\sin(22^\circ) \cdot \sin(22.5^\circ)}{\sin(0.5^\circ)} \]

### Step 3: Substitute the sums into the expression for x
Now substituting these results back into the expression for \(x\):

\[ x = \frac{\frac{\sin(22^\circ) \cdot \cos(22.5^\circ)}{\sin(0.5^\circ)}}{\frac{\sin(22^\circ) \cdot \sin(22.5^\circ)}{\sin(0.5^\circ)}} \]

The \(\sin(22^\circ)\) and \(\sin(0.5^\circ)\) cancel out:

\[ x = \frac{\cos(22.5^\circ)}{\sin(22.5^\circ)} = \cot(22.5^\circ) \]

### Step 4: Calculate the value of cot(22.5°)
Using the identity:

\[ \cot(22.5^\circ) = \sqrt{2} + 1 \]

### Step 5: Find the greatest integer less than or equal to x
Now, we need to find the greatest integer that does not exceed \(x\):

\[ x = \sqrt{2} + 1 \approx 1.414 + 1 = 2.414 \]

Thus, the greatest integer that does not exceed \(x\) is:

\[ \lfloor x \rfloor = 2 \]

### Final Answer
The greatest integer that does not exceed \(x\) is \(2\).