How should I start?

$\begin{aligned} S & = \frac 1{90} \sum_{n=1}^{90} 2n \sin (2n^\circ) \\ & = \frac 1{90} \left( \color{#3D99F6}{2\sin 2^\circ} + \color{#20A900}{4\sin 4^\circ} + \color{#EC7300}{6\sin 6^\circ} + \cdots + \color{#EC7300}{174\sin 174^\circ} + \color{#20A900}{176\sin 176^\circ} + \color{#3D99F6}{178\sin 178^\circ} + \color{#D61F06}{180\sin 180^\circ} \right) \\ & = \frac 1{90} \left( \color{#3D99F6}{2\sin 2^\circ} + \color{#20A900}{4\sin 4^\circ} + \color{#EC7300}{6\sin 6^\circ} + \cdots + \color{#EC7300}{174\sin 174^\circ} + \color{#20A900}{176\sin 176^\circ} + \color{#3D99F6}{178\sin 178^\circ} + \color{#D61F06}{0} \right) \end{aligned}$

Note that:

$\begin{aligned} \sin (180^\circ - \theta) & = \sin \theta & \small \color{#EC7300}{\text{For example: }\sin 6^\circ =\sin 174^\circ} \\ \implies 2n \sin (2n^\circ) + (180-2n) \sin (180^\circ - 2n^\circ) & = 180 \sin 2n^\circ = 180 \sin (180^\circ - 2n^\circ) \\ & = 90 \sin 2n^\circ + 90 \sin (180^\circ - 2n^\circ) \end{aligned}$

Therefore, we have:

$\begin{aligned} S & = \frac 1{90} \left(90 \sin 2^\circ + 90 \sin 4^\circ + 90 \sin 6^\circ + \cdots + 90 \sin 90^\circ + \cdots + 90 \sin 174^\circ + 90 \sin 176^\circ + 90 \sin 178^\circ \right) \\ & = \color{#3D99F6}{\sum_{n=1}^{89} \sin (2n^\circ) = \sum_{n=1}^{89} \sin \left( \frac {n\pi}{90} \right) = \cot \left(\frac \pi{180} \right) = \cot 1^\circ \quad \quad \small \text{See Note.}} \end{aligned}$

$\implies k = \boxed{1}$

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$\color{#3D99F6}{\text{Note:}} \quad \displaystyle \sum_{n=1}^{m-1} \sin \left(\frac {n \pi}m \right) = \cot \left(\frac \pi{2m} \right) \quad$ (T. Drane, pers. comm., Apr. 19, 2006 -- Eqn. 21) .