Answer = $C.$
We will make the pairs of $\sin^2\theta + \cos^2\theta$, where $\theta$ is ame for both $\sin$ and $\cos$
$$\sin^25^0 + \sin^285^0 = \sin^25^0 + \sin^2(90^0-5^0) = \sin^25^0 + \cos^25^0 =1$$
(Using $\color{blue}{\sin^2\theta + \cos^2\theta = 1}$)
$$\sin^2{85}^0 = \cos^25^0 (\color{blue}{\because \sin\theta = \cos({90^0-\theta})})$$
similarly,
$$\sin^210 \;and\; \sin^280$$
$$\vdots$$
$$\sin^245^0 = \bigg(\frac{1}{\sqrt2}\bigg)^2 = \frac{1}{2} = 0.5$$
$$\vdots$$
$$\sin^240^0 \;and\; \sin^250$$
$$\vdots$$
$$\sin^290^0 = (1)^2 = 1$$
So, this way $8$ pairs will be formed, leaving $\sin^245^0$ and $\sin^290^0$
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 $