Find the volume of a triangular pyramid which has base with edges of length 4, 5 and 7, and lateral faces with the base form a 48 degrees and 30 minutes angle. Solution is volume $V=4.52$; base $B=4\sqrt{6}$.
What I am interested about is how the height $h$ of that tetrahedron was found. I don't have the whole solution just meta results so I reconstructed how it was calculated.
In the text it doesn't say whether it's right or oblique pyramid. It was calculated using the radius of an inscribed circle in the base $r$, $r=\frac{B}{s}$, $s=\frac{a+b+c}{2}$. Then look at the triangle that is made out of the height of the tetrahedron $h$, radius of the inscribed circle in the base $r$ and slant height of one of the lateral faces $s$. It is assumed that that is a right triangle. That allows to calculate the size of the 3rd angle in that triangle, which is 41 degrees and 30 minutes. So using the law of sines ($\frac{h}{\sin48.5}=\frac{r}{\sin41.5}$) the height $h$ was found.
The question is how do we know that the triangle used in the calculation is a right triangle, how do we know that the foot of the height $h$ of that tetrahedron coincides with the center of the circle that is inscribed in the base and has radius $r$? How to be sure whether it's right or oblique pyramid?
Call $V$ the vertex of the pyramid and consider its foot $H$. From $H$, draw the normal to one side of the base, and call $A$ the point in which this normal crosses the side. The triangle $VHA$ is right, because $VH$ and $HA$ are perpendicular. Also, we have $\angle{HAV}=48°30'$.
Now repeat the same procedure with the other two sides, and call $B$ and $C$ the points in which the normals cross the sides. In this way, we get the two new triangles $VHB$ and $VHC$. Since these triangles are right as well, and because $\angle{HBV}=48°30'$ and $\angle{HCV}=48°30'$, the three triangles have equal angles (for all the triangles, the third angle, which is at the vertex, is $90°-48°30'=41°30'$).
Now it is sufficient to note the three triangles also have the side $VH$ in common, so they are equal. Therefore $HA=HB=HC$, which proves that the foot of the height of the tetrahedron coincides with the center of the circle that is inscribed in the base, i.e. the pyramid is right.