I'm a little baffled by the following pentagonal tiling:
It clearly has an 8-tile primitive unit, and thus should be one of types 7-8 (Richard Kershner) or else one of types 9, 11, 12 or 13 (Marjorie Rice) in
https://en.wikipedia.org/wiki/Pentagonal_tiling
However, the angles (as appearing in the photograph, or as in the diagrams below) do not seem to match any of these types (though perhaps, indeed probably, I'm making a mistake). What gives? What is the type of this tiling?
(Also, what are the angles, exactly? And are all five side of exactly the same length?)
I think it's one form of Type 8. It may be a deformation of the equilateral form of the Cairo pentagonal tiling, but Type 8 seems to fit.
Go to the Wolfram demo on pentagonal tilings and select Type 8. There appear to be three parameters in type 8, but the demo allows you to vary only one parameter. The default setting yields a tiling that corresponds closely to the image you've posted:
Alternatively, in the Wikipedia page on pentagonal tilings, you can stare at the gif in the table that defines Type 8, and eventually it cycles to a form similar to the one in your image.
ADDED: The pentagons that form Type 8 tilings have four equal sides and the angles satisfy two constraints. Adopting the labeling found in the article Tiling the Plane with Congruent Pentagons by Doris Schattschneider, the pentagons are characterized by $$ 2A+B=2\pi, \quad 2D+C=2\pi, \quad a=b=c=d.\tag1 $$ Once you've specified a value for $B$ (say), the remaining angles are determined by the above constraints, so there's really only one degree of freedom for a given value of $a$, which explains why there is only one active control in the Wolfram demo for Type 8.
There is a unique Type 8 tiling where all sides are equal. For this equilateral pentagon the Schattschneider article derives the angles (see Table III, Type 8): $$E=\arccos \frac {\sqrt{13}-3}4, \quad B=C=\pi-E, \quad A=D=\frac\pi2 +\frac E2,$$ or $$A=D\approx 130^\circ39',\quad B=C\approx 98^\circ 42', \quad E\approx 81^\circ 18'.$$ In the Wolfram demo on pentagonal tilings, set the parameter AA, which corresponds to angle $B$, to $1.72$ (radians). This yields the equilateral Type 8 tiling. The fact that $B+E=C+E=\pi$ means that hexagons will be seen in the result. In your tiling the hexagons are not quite there, so your tiling is described by a slightly different value for AA. (The parameter AA apparently corresponds to the angle formed by the horizontal segment seen at the very center of the demo; it's slightly less than $90^\circ$ in the above screenshot.)
ADDED 2: Given a desired value for angle $B$ or non-equal side $e$, you can solve the rest of the Type 8 pentagon by combining constraints (1) with the law of cosines for pentagons: $$ a^2+b^2+c^2 - 2ab\cos A - 2bc \cos B + 2ac \cos(A+B)=d^2+e^2-2de\cos D.$$ As an example, this Wolfram Alpha query sets $a=b=c=d=1$ and $e=x=1$ to solve the equilateral pentagon: $$A\approx 2.28021, \quad B≈1.72277, \quad C\approx 1.72277, \quad D\approx 2.28021,\quad E\approx 1.41882 $$