What is the sum of the angles in a right triangle?

[anemone]

Here is this week's POTW:

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Given that \(\displaystyle \sum_{k=1}^{35} \sin 5k=\tan \dfrac{a}{b}\), where angles are measured in degrees and $a$ and $b$ are relatively prime positive integers that satisfy $\dfrac{a}{b}<90$, evaluate $a+b$.

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[anemone]

Congratulations to the following members for their correct solution: (Smile)

1. castor28
2. lfdahl
3. greg1313
4. kaliprasad

Solution from lfdahl:

Spoiler

By Lagrange´s trigonometric identities (angles in degrees), we get

\[\sum_{k=1}^{35}\sin (5k) = \frac{1}{2}\cot \left ( \frac{5}{2} \right )-\frac{\cos \left ( 175 + \frac{5}{2} \right )}{\sin \left ( \frac{5}{2} \right )} = \tan \left ( \frac{175}{2} \right )\]

Since $2$ and $175$ are coprimes and $\frac{175}{2} < 90$, we can conclude that $a + b = 177$.

Alternate solution from castor28:

Spoiler

Let us write $\alpha=5\mbox{°}=\pi/36$ and $\theta=e^{i\alpha}=\cos\alpha+i\cdot\sin\alpha$.
The sum is the imaginary part of $\displaystyle S = \sum_{k=1}^{35}{\theta^k}$. (Actually, because of the symmetry of the pattern with respect to the $y$-axis, $S$ is pure imaginary, and we only need $S/i$).
As the terms of $S$ constitute a geometric progression, we have:

$$\begin{align*}
S &= \frac{\theta^{36}-\theta}{\theta-1}\\
&= \frac{-1-e^{i\alpha}}{e^{i\alpha}-1}\quad\text{because }\theta^{36}=e^{i\pi}=-1\\
&= -\frac{e^{i\alpha/2}+e^{-i\alpha/2}}{e^{i\alpha/2}-e^{-i\alpha/2}}\\
&= i\cot\alpha/2 = i\cot2.5\mbox{°}\\
&= i\tan87.5\mbox{°}
\end{align*}$$
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This gives $a=175$, $b=2$ and $a+b=177$