This is a challenging one: Given all the information at one corner of a tetrahedron (all three surface angles and all three edge lengths), what is the volume of the tetrahedron?

The volume of any pyramid is equal to the area of its base times its height, divided by three: \[V = \frac{Ah}{3}\]

Since a tetrahedron is a pyramid from all of its faces, we would expect the final formula to be symmetrical. I’ll use \(\Delta BCD\) as the base, but we should come up with the same formula using \(\Delta ACD\) or \(\Delta ABD\).

The area of a triangle given two sides and the shared angle is given by \[A = \frac{bc \sin \beta}{2}\]

This can be easily seen by drawing \(\overline{BZ} \perp \overline{CD}\). The area of \(\Delta BCD\) is \(\frac{b \cdot BZ}{2}\); \(\sin \beta = \frac{BZ}{c}\), so \(BZ = c \sin\beta\) and thus \(A = \frac{bc \sin \beta}{2}\).

Next, calculate the height of this tetrahedron. I’ll draw in two line segments. \(h \, (\overline{EA})\) is the height, and is perpendicular to \(\Delta BCD\). \(\Delta AED\) is a right triangle, so \(h^2 + d^2 = a^2\) and \(h^2 = a^2 – d^2\). We are given \(a\), so if we can calculate \(d\), we can calculate \(h\).

Consider the base, \(\Delta BCD\). I’ve added three more segments: \(\overline{EG}\perp\overline{CD}\), \(\overline{EF}\perp\overline{BD}\), and \(g \, (\overline{FG})\). This gives us the quadrilateral \(EFDG\), where \(\angle F\) and \(\angle G\) are right angles. Because we have a quadrilateral with opposite angles that are supplementary, we know that the four vertices fall on a circle.

In the next diagram, I’ve drawn the circle around the quadrilateral and added the center point and two radii. Because \(\angle FDG\) is an inscribed angle, it is half the size of central angle \(\angle FHG\). By the law of cosines, \(g^2 = r^2 + r^2 – 2\cdot r \cdot r \cdot \cos{2\beta} = 2r^2 (1 – \cos{2\beta})\). We use the trigonometric identity \(\cos{2\beta} = 1 – 2\sin^2\beta\) to determine that \(g^2 = 4r^2\sin^2\beta\) and \(r = \frac{g}{2\sin\beta}\). Thus, \(d = 2r = \frac{g}{\sin\beta}\).

From the law of cosines, we know that \(g^2 = DG^2 + DF^2 – 2 \cdot DG \cdot DF \cdot \cos\beta\).

Since \(\Delta AFD\) is a right triangle, \(DF = a \cos\alpha\). Likewise, \(DG = a \cos\gamma\). Hence, \(g^2 = a^2 \cos^2\alpha + a^2 \cos^2\gamma – 2 \cdot a \cos\alpha \cdot a \cos\gamma \cdot \cos\beta\) \(=a^2 (\cos^2\alpha + \cos^2\gamma – 2\cos\alpha\cos\beta\cos\gamma)\).

We can now calculate \(h\). \(d^2 = \frac{g^2}{\sin^2\beta} = \frac{a^2 (\cos^2\alpha + \cos^2\gamma – 2\cos\alpha\cos\beta\cos\gamma)}{\sin^2\beta}\), so \(h^2 = a^2 – \frac{a^2 (\cos^2\alpha + \cos^2\gamma – 2\cos\alpha\cos\beta\cos\gamma)}{\sin^2\beta}\) \(= \frac{a^2(\sin^2\beta – \cos^2\alpha – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma)}{\sin^2\beta}\) \(=\frac{a^2(1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma)}{\sin^2\beta}\). Thus, \(h = \frac{a}{\sin\beta}\sqrt{1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma}\).

Finally, \(V = \frac{Ah}{3} = \frac{b\cdot c\cdot \sin\beta \cdot a}{3 \cdot 2 \cdot \sin\beta} \sqrt{1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma}\) \(=\frac{abc}{6} \sqrt{1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma}\).

So, given the three surface angles and three edge lengths from a given vertex of a tetrahedron, the volume of the tetrahedron is:

\[V=\frac{abc}{6} \sqrt{1-\cos^2\alpha – \cos^2\beta – \cos^2\gamma + 2\cos\alpha\cos\beta\cos\gamma}\]

Very good! Congrats!