Deep Pit

To use formulas for sphere inscribed in a solid, we need that solid. So let’s add a fourth plane, and for simplicity make it perpendicular to line $AB$ , that is the edge between the $10^\circ$ face and the $15^\circ$ face. And let's for now set the distance $AB=1$ .

To calculate $x$ , we’ll use law of cosines in the triangle $ACD$ with sides $AC=\frac{1}{cos(10^\circ)}$ and $AD=\frac{1}{cos(15^\circ)}$ and angle between them of $20^\circ$ . It should come to about $x=0.3566$ .

Law of cosines applied to the triangle $DBC$ with sides $DB=tan(15^\circ), BC=tan(10^\circ)$ , and $x$ will give us the angle $DBC=104.9^\circ$ . From that the area of the top will be $A_{top}=0.0228$ .

Areas of the other faces are easy. Area of $\triangle ABC=\frac{1}{2}tan(10^\circ)$ . Area of $\triangle ABD=\frac{1}{2}tan(15^\circ)$ .

Area of $\triangle ACD=\frac{1}{2}\times\frac{1}{cos(10^\circ)}\times \frac{1}{cos(10^\circ)}\times sin(20^\circ)$ . And the volume of the irregular tetrahedron is $V=\frac{1}{3}\times 1\times A_{top}$

We can get the radius of the inscribed sphere from $V=\frac{1}{3}\times R\times \sum {A_i}$ and it comes to $R=0.0537$ .

The problem states that it should be 10cm, so everything needs to be scaled accordingly and the distance $AB$ instead of being 1 will be $AB\approx 186.055$ .

Now the job is to find the distance from the sphere to point $A$ . Thanks to the two right angles at $B$ , calculating the distance from center of the sphere $O$ to line $AB$ is easy.

In the top view on the left we have a right triangle $OTS$ with angle $OST=\frac{\angle DBC}{2}=52.45^\circ$ and a leg $OT=R=10$ , so $OS=\frac{10}{sin(52.45^\circ)}=12.61$ .

In the side view on the right we’ll use Pythagorean theorem in the triangle $ASO$ with right angle at $S$ . The other leg is $AS=AB-10$ . So the hypotenuse is $176.5$ and the distance between the sphere and point $A$ is $166.5$ centimeters.