Given that `sum_(k=1)^35 sin5k^@ = tan(m/n)^@` , where m and n are relatively prime positive integers that satisfy `(m/n)^@<90^@`, then m + n is equal to

Options

A. 173
B. 175
C. 177
D. 179

Answer

To solve the problem, we need to evaluate the sum \( S = \sum_{k=1}^{35} \sin(5k^\circ) \) and express it in the form \( \tan\left(\frac{m}{n}\right)^\circ \), where \( m \) and \( n \) are relatively prime positive integers.

### Step-by-Step Solution:

1. **Define the Sum**:
Let \( S = \sum_{k=1}^{35} \sin(5k^\circ) \).

2. **Use the Sine Sum Formula**:
We can use the formula for the sum of sines:
\[ \sum_{k=1}^{n} \sin(a + kd) = \frac{\sin\left(\frac{nd}{2}\right) \sin\left(a + \frac{(n+1)d}{2}\right)}{\sin\left(\frac{d}{2}\right)} \]
Here, \( a = 5^\circ \), \( d = 5^\circ \), and \( n = 35 \).

3. **Apply the Formula**:
Plugging in the values:
\[ S = \frac{\sin\left(\frac{35 \cdot 5^\circ}{2}\right) \sin\left(5^\circ + \frac{(35+1) \cdot 5^\circ}{2}\right)}{\sin\left(\frac{5^\circ}{2}\right)} \]
Simplifying:
\[ S = \frac{\sin(87.5^\circ) \sin(90^\circ)}{\sin(2.5^\circ)} \]
Since \( \sin(90^\circ) = 1 \):
\[ S = \frac{\sin(87.5^\circ)}{\sin(2.5^\circ)} \]

4. **Use the Identity**:
We know that \( \sin(87.5^\circ) = \cos(2.5^\circ) \):
\[ S = \frac{\cos(2.5^\circ)}{\sin(2.5^\circ)} = \cot(2.5^\circ) \]

5. **Express in Terms of Tangent**:
Since \( \cot(x) = \frac{1}{\tan(x)} \):
\[ S = \cot(2.5^\circ) = \tan(90^\circ - 2.5^\circ) = \tan(87.5^\circ) \]

6. **Identify m and n**:
Now, we can express \( S \) as:
\[ S = \tan\left(\frac{175}{2}\right)^\circ \]
Here, \( m = 175 \) and \( n = 2 \).

7. **Check for Relative Primality**:
The integers \( 175 \) and \( 2 \) are relatively prime.

8. **Calculate m + n**:
Finally, we find:
\[ m + n = 175 + 2 = 177 \]

### Conclusion:
Thus, the final answer is:
\[ \boxed{177} \]