Mangle Angle Triangle

Let $D$ be the point of tangency on $AB$ , $F$ be the point of tangency on $AC$ , and $O$ be the center of the incircle, and let $a = CF = CE$ , $b = BE = BD$ , and $c = AF = AD$ .

Then the area of the triangle is given by $40 = \frac{1}{2}(a + c)(b + c) \sin 75°$ , which rearranges to $ab + ac + bc + c^2 = 80 \csc 75°$ .

From $\triangle OAD$ , the inradius is $r = c \tan 37.5°$ , but also $r = \cfrac{A_{\triangle ABC}}{s_{\triangle ABC}} = \cfrac{40}{a + b + c}$ , so $c \tan 37.5° = \cfrac{40}{a + b + c}$ , which rearranges to $ac + bc + c^2 = 40 \cot 37.5°$ .

Combining $ab + ac + bc + c^2 = 80 \csc 75°$ and $ac + bc + c^2 = 40 \cot 37.5°$ gives $CE \cdot EB = ab = 80 \csc 75° - 40 \cot 37.5° \approx 30.6930795191584$ , making the sum of the first $11$ digits to the right of the decimal point $6 + 9 + 3 + 0 + 7 + 9 + 5 + 1 + 9 + 1 + 5 = \boxed{55}$ .