please, how to solve this particularsangaku?
Here, sun is a japanese measure unity.
sangakus are part of the japanese tradition and are of interest of math problem solvers and enhusiasts and i've not found a solution, the sangaku is taken from this PhD Thesis:https://www.academia.edu/30061089/_PhD_Thesis_Sangaku_A_Mathematical_Artistic_Religious_and_Diagrammatic_Examination.
this particular sangaku is important for me because this it is a sangaku with a photograph presenting typical sangakus sttructure(please see the image)and it is a pity not having its solution
It's not entirely clear in the diagram, but I'm taking the lower two triangles to have collinear sides, and the upper to triangles to have perpendicular sides. (Luckily, the consequence of these assumptions matches the numerical value given in the problem.)
Let the bounding circle have center $O$ and radius $r$; let the triangles have side-length $2s$; let the upper triangles make a right angle at $D$, and define $d:=|OD|$; and let the target triangle have centers $P$ and $Q$ and radii $p$ and $q$, as shown.
Note that $P$ and $Q$ line on the bisector of the angles made by triangle sides at $D$.
From the figure at left above, we deduce without too much trouble that $$r^2 = s^2\left(1^2+(\sqrt{3}+1)\right)^2=s^2(5+2\sqrt{3})$$ $$d^2 = s^2\left(1^2+(\sqrt{3}-1)\right)^2=s^2(5-2\sqrt{3})$$ $$\cos\angle ODE = -\frac{s}{d}(\sqrt{3}-1) \qquad \sin\angle ODE = \frac{s}{d}$$ Defining $\theta := \angle ODP = \angle ODE-105^\circ$ we can calculate $$\cos\theta := \cos\angle ODE \cos 105^\circ + \sin\angle ODE \sin 105^\circ = \frac{s}{d}\cdot\frac{5-\sqrt{3}}{2\sqrt{2}}$$ The tangencies of $\bigcirc P$ and $\bigcirc Q$ imply $$\begin{align} |OP| &= r-p \qquad |DP| = p\csc45^\circ \\ |OQ| &= r-q \qquad |DQ| = q\csc75^\circ \end{align}$$ so that applying the Law of Cosines to $\triangle ODP$ and $\triangle ODQ$ gives $$\begin{align} (r-p)^2 &= d^2 + p^2\csc^245^\circ - 2 d p \csc45^\circ \cos\theta \\ (r-q)^2 &= d^2 + q^2\csc^275^\circ + 2 d q \csc75^\circ \cos\theta \end{align}$$ Solving these quadratics for $p$ and $q$ gives these positive roots: $$\begin{align} \frac{p}{s} &= \frac12 \left( \mu - 2\lambda + \sqrt{2 (24 + 7\sqrt3 - 2\lambda\mu)} \right) && = 1.649\ldots \\[4pt] \frac{q}{s} &= (7+\sqrt{3})\left( \nu - \lambda + \sqrt{ 2 (3 \sqrt3 - \lambda\nu)} \right) && = 0.837\ldots \\[4pt] \lambda &:= \sqrt{5+2\sqrt{3}} \qquad \mu := 5-\sqrt{3} \qquad \nu := 4-3\sqrt{3} \end{align}$$
Thus, $q/p = 0.507\ldots$, so that for $p=12$, we have $q = 6.0937\ldots$. $\square$
If there are nicer forms of $p$ and $q$, then I'm not sure what they are. And yet, I suspect additional structure lurks in the configuration, since the legs of the right triangle with hypotenuse $d$ are $$s = \frac{s\csc45^\circ}{\sqrt2} \qquad s(\sqrt{3}-1) = \frac{s\csc75^\circ}{\sqrt2}$$