2024 AMC 12B Problem 11

Using $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2},$ $$\begin{aligned} \sum_{n=1}^{90} \sin^2(n^\circ) &= \frac{90}{2} \\ &\quad {}- \frac12 \sum_{n=1}^{90}\cos(2n^\circ). \end{aligned}$$ In the cosine sum, the terms for $n$ and $90 - n$ satisfy $\cos(2n^\circ) + \cos(180^\circ - 2n^\circ) = 0,$ and $\cos 90^\circ = 0,$ so everything cancels except $\cos 180^\circ = -1.$ Hence the sum is $45 - \tfrac12(-1) = 45.5,$ and the mean is $\dfrac{45.5}{90} = \dfrac{91}{180}.$ Thus, the correct answer is E.