Why reflected light from a mirror don’t get absorbed in polarized sunglasses? If i see a reflection on a horizontal surface and i try to cancel it out using my polarized sunglasses, then it cancels out much of that reflection. But when i put a small mirror inside that reflection on the same surface and try to cancel out the reflection from that surface using polarized sunglasses, then it cancels out much of that reflection, but it do NOT cancel out the light that is reflected from the surface of the mirror. Why is that?
Please refer the image linked below, if you can not understand what i am describing

 A: Polarization results when unpolarized light impinges on a transparent medium at (or near) Brewster's angle :

A mirror (electrically conducting surface) reflects both polarizations equally.
Your "horizontal surface" may be something like smooth "shiny" "black" plastic, but if it is "black" then why does it reflect at all? Because it is a  dielectric material with an index of refraction different from air that has light absorbing particles in it. On the scale of a few light wavelengths  it acts like a transparent medium.
A: Brewster's angle is only a special case and not as crucial as people might think in this case. To keep things short:
Mirror: Conventional mirrors have a metallic surface (e.g. aluminum or silver) where the mobile surface charge carriers cause a reflection of the light. Polarization usually does not matter enough here to cause a significant polarization of the reflected light. Hence, it can only be partially blocked by a polarization filter. You can look at Graphs for s- and p-polarization on this page (click on "Graphs" for plots) to verify that s- and p-pol don't make too much of a difference here.
Floor: Dielectric (non-conducting) materials also reflect a part of the light according to the Fresnel equations (This page actually shows a figure related to your question). For large angles of incidence (let's say roughly larger than $40°$) the reflection properties are usually completely different for s- and p-pol. P-pol light is then reflected poorly while a part of the s-pol light still gets reflected. Brewster's angle is an extreme case of this ($0\%$ reflection of p-pol light), but generally the p-pol reflection is much weaker for steep angles. This causes a strong polarization of the reflection.
