Approximate the answer based on distances between SSS triangles' upper vertices.

Let us label the angles as such. Using cosine rule , we have

$\cos(\theta_1) = \dfrac{d^2 + 23^2 - 27^2}{2\cdot d \cdot 23}, \quad \cos(\theta_2) = \dfrac{d^2 + 16^2-22^2}{2\cdot d \cdot 16}.$

With compound angle formula , $\cos(A+B) = \cos(A) \cos(B) - \sin(A) \sin(B)$ , we obtain

$\cos(360^\circ - (\theta_1 + \theta_2)) = \cos(\theta_1 + \theta_2) = \cos(\theta_1) \cos(\theta_2) - \sin(\theta_1) \sin(\theta_2) = \dfrac{23^2 + 16^2 - 30^2}{2\cdot23\cdot16} = -\frac5{32}.$

Because $\sin(A) = \sqrt{1-\cos^2(A)}$ for $0^\circ < A < 180^\circ$ , the equation above can be expanded and simplified to:

$900 d^4 - \num{921315} d^2 + \num{42983536}= 0 .$ Using quadratic formula , and knowing that $d>0$ is super duper close to an integer, $d = \frac12\sqrt{\dfrac{\num{61421}}{30} - \dfrac{23}2 \sqrt{\dfrac{\num{388759}}{15} }} \approx 7 + \boxed{0.00000008573675}.$

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Also note that the inradius of the circumcircle of the entire triangle is $\dfrac{ \sqrt{s(s-a)(s-b)(s-c)}}s$ , where $a,b,c = 27,22,30, s = \frac{a+b+c}2$ , then $r = \sqrt{ \frac{\num{16625}}{316}} < 8$ . So $d < 2r < 16$ ( $d$ is shorter than the diameter of the circumcircle). Thus, even if we're not told that $d$ is super duper close to an integer, then the other positive root of $d$ (in the quartic equation above) is an extraneous root.

On the other hand, by trial and error, $900 X^4 - \num{921315} X^2 + \num{42983535 } = 0$ has a root 7, so $900 X^4 - \num{921315} X^2 + \num{42983535}+ 1=0$ has a root larger than 7, but only eensy larger than 7.

I solved this problem using GeoGebra - check out the link below to see what I did.

https://www.geogebra.org/calculator/tjbwrbrg

:-) I had more problems generating the illustration than doing the problem.

Assuming that the base of both triangles is horizontal and that the left hand ends of the bases are at the origin, the coordinates of the vertices of the larger triangle are $\{0,0\},\{30,0\},\left\{\frac{229}{12},\frac{\sqrt{52535}}{12}\right\}$ and the coordinates of the vertices of the smalle triangle are $\{0,0\},\{30,0\},\left\{\frac{391}{20},\frac{23 \sqrt{111}}{20}\right\}$ . The distance $d$ is $\sqrt{\left(\frac{\sqrt{52535}}{12}-\frac{23 \sqrt{111}}{20}\right)^2+\frac{49}{225}}$ . That value evaluates after rounding to $7.$ The difference numerically between $7$ and $d$ is about $\text{8.573674836753753E-8}$ .