A right triangle has a hypotenuse of length 40 and an angle of 25°, with a side opposite this angle of length 16. A second right triangle also has an angle of 25°, with a hypotenuse of length 10. Determine the length of the side opposite the 25° angle on the second triangle.
We begin with a diagram: [asy]
pair X=(0,0), Y=(9.3235,0), Z=(0,25.125), M=(9.3235,2.641);
draw(X--Y--Z--X);
label("$A$",X,SW);
label("$B$",Y,SE);
label("$C$",Z,N);
draw(rightanglemark(Y,X,Z,40));
label("$D$",M,NE);
draw(M--Y);
[/asy] Let $A,B$ and $C$ be points such that $AB=13.446,$ $BC=40,$ and $\angle ABC=90^\circ.$ Suppose that $C$ and $D$ are on opposite sides of line $AB.$ Put triangle $ACD$ in the Euclidean plane and unfold it so that the figure coincides with the image below: [asy]
pair X=(0,0), Y=(9.3235,0), Z=(0,25.125), M=(9.3235,2.641);
draw(X--Y--Z--X);
draw(M--Y);
draw((4.6617,50.250)--(4.6617,0));
draw((0,50.250)--X--Y);
label("$A$",X,SW);
label("$B$",Y,SE);
label("$C$",Z,N);
label("$D$",M,NE);
label("$D$",(4.6617,50.250),N);
label("$C$",(0,50.250),N);
label("$D$",(4.6617,-1.5),S);
label("$C$",(-1.75,0),W);
[/asy] Note that $BD=40-16=24.$ We begin by performing an angle bisector construction: [asy]
pair X=(0,0), Y=(9.3235,0), Z=(0,25.125), M=(9.3235,2.641);
draw(X--Y--Z--X);
draw(M--Y);
draw((4.6617,50.250)--(4.6617,0));
draw((0,50.250)--X--Y);
draw(Z--M--X,solid+dashed);
label("$A$",X,SW);
label("$B$",Y,SE);
label("$C$",Z,N);
label("$D$",M,NE);
label("$D$",(4.6617,50.250),N);
label("$C$",(0,50.250),N);
label("$D$",(4.6617,-1.5),S);
label("$C$",(-1.75,0),W);
label("$E$",X,SW);
[/asy] Let $E$ be the intersection of $\overline{AC}$ and $\overline{BD}.$ Using the Angle Bisector Theorem on $\triangle ABC,$ we have $$\frac{BC}{AB}=\frac{CD}{AD}\implies\frac{40}{13.446}=\frac{CD}{DA}\implies CD=3.908{{4}{9}}.$$ We now apply the Law of Cosines on $\triangle ABC:$ $$BC^2=AB^2+AC^2-2(AB)(AC)\cos\angle BAC\implies40^2=13.446^2+AC^2-2(13.446)(AC)\cos(90^\circ-25^\circ).$$ Simplifying, we have $AC^2\approx477.48.$ By the Distance Formula on $\triangle ADC$ and $\triangle AEC,$ respectively, we obtain the following equalities: \begin{align*}
CD^2+AD^2&=AC^2,\\
CE^2+AE^2&=AC^2.
\end{align*} Substituting known values into each equation and solving each for the desired length yields \begin{align*}
3.9084^2+AD^2&\approx477.48,\\
CE^2+16^2&\approx477.48.
\end{align*} Subtracting the latter equation from the former and simplifying gives $AD\approx 4.2839.$ Since $\triangle ADE\sim\triangle BDC,$ it follows that $$\frac{AD}{BD}=\frac{DE}{CD}\implies DE\approx8.33.$$ The Law of Cosines on $\triangle ADE$ gives $$8.33^2=4.2839^2+DE^2-2(4.2839)(DE)\cos\angle DAE\approx16.676+68.68-67.968.$$ Solving gives $\cos\angle DAE\approx0.343,$ so $\sin\angle DAE\approx\sqrt{1-\cos^2\angle DAE}\approx0.94.$ Hence, we have $$\sin\angle ADE=\frac{AD}{DE}\sin\angle DAE\approx\frac{4.2839}{8.33}(0.94)\approx 0.484.$$ Applying the Law of Sines on $\triangle ADE$ gives $$\frac{16}{\sin\angle ADE}=\frac{8.33}{\sin\angle AED},$$ so $\sin\angle AED\approx\tfrac{8.33}{16}(0.484)\approx0.25.$ Hence, $$\sin\angle ADB=\frac{BD}{AB}\sin\angle AED=\frac{24}{13.446}(0.25)\approx 0.441.$$ Since $\angle ADB<90^\circ,$ we have $\cos\angle ADB\approx0.897,$ so $\angle ADB\approx 26.44^\circ.$ Finally, we apply the Law of Sines on $\triangle ABD$ to find the third angle for $\triangle ABD,$ from which we can deduce that the side opposite the $25^\circ$ angle on $\triangle ABD$ has length \begin{align*}
\frac{16}{\sin\angle ADB}&=\frac{BD}{\sin\angle BAD}\\
\implies\frac{16}{\sin 26.44^\circ}&=\frac{24}{\sin\angle BAD}\\
\implies\frac{1}{\sin 26.44^\circ}&=\frac{3}{\sin\angle BAD}\\
\implies\sin\angle BAD&\approx\frac{3}{\sin 26.44^\circ}\approx 1.72\\
\implies\angle BAD&\approx 63.9^\circ\\
\implies AD&\approx\frac{\sin(90^\circ-63.9^\circ)}{\sin 63.9^\circ}\cdot 24\approx 17.25.\end{align*}
[asy]
unitsize(1inch);
pair A,B,C,D,E;
A=(0,0);
B=(13.446,0);
C=(0,40);
D=IP(Circle(A,4.29),arc(B,A,D));
E=(4.29,0);
label("$A$",A,(SW));
label("$B$",B,(S));
label("$C$",C,(NE));
label("$D$",D,(NW));
label("$E$",E,(S));
draw(A--B--C--cycle);
draw(D--B,dashed);
label("$24$",(E--B),south);
label("$13.446$",(A--B),south);
label("$40$",(B--C),NE);
label("$3.9084$",(C--D),NE);
label("$16$",(A--E),south);
[/asy] Checking a super-accurate diagram, we see that the side opposite the $25^\circ$ angle on the second triangle looks to be a bit above $3.5.$ Thus, the answer is $\boxed{3.516}.$[asy]
unitsize(1inch);
pair A,B,C,D,E;
A=(0,0);
B=(13.446,0);
C=(0,40);
D=IP(Circle(A,4.29),arc(B,A,D));
E=(4.29,0);
A=(0,0);
B=(13.446,0);
C=(0,40);
D=IP(Circle(A,4.29),arc(B,A,D));
E=(4.29,0);
label("$A$",A,(SW));
label("$B$",B,(S));
label("$C$",C,(NE));
draw(A--B--C--cycle);[/asy]