$\underline{\textbf{Answer:C}\Rightarrow}$
$\underline{\textbf{Solution:}\Rightarrow}$
We will make the pairs of $\sin^2\theta + \cos^2\theta$, where $\theta$ is ame for both $\sin$ and $\cos$.
$\sin^25^0 + \sin^285^0 = \sin^25^0 + \sin^2(90^0-5^0) = \sin^25^0 + \cos^25^0 =1\;\;\;\mathrm{[Using\;\;sin^2\theta + \cos^2\theta = 1]}$
$\sin^2{85}^0 = \cos^25^0 (\color{blue}{\because \sin\theta = \cos({90^0-\theta})})$
$\text{Similarly,}\;\;\;\sin^210 \;\text{and}\; \sin^280$
$\vdots$
$\sin^245^0 = \bigg(\frac{1}{\sqrt2}\bigg)^2 = \frac{1}{2} = 0.5$
$\vdots$
$\sin^240^0 \;\text{and}\; \sin^250$
$\vdots$
$\sin^290^0 = (1)^2 = 1$
So, this way $\mathbf 8$ pairs will be formed, leaving $\sin^245^0$ and $\sin^290^0$
$\text{But we know that:}\;\;\;\sin^245^0 = \bigg(\frac{1}{\sqrt2}\bigg)^2 = \frac{1}{2} = 0.5$
$\sin^290^0 = 1^2 = 1$
So, answer $= 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 +0.5 = 8\times1 +0.5 =9.5 $
$\therefore \mathbf C$ is the right option.
