Thursday, 1 October 2009

Heron's Formula, The Trigonometric Version

Here is a nice problem to challenge your geometry teacher with. Ask her/him to find the volume of a parallelepiped (think of a box except the sides are not right angles... a prism with six faces, all parallelograms.) Probably they will say, "No problem, I'll find the base area and multiply by the height." Or maybe he is a vector fan and would prefer to use the scaler triple product. But which of those do you try if you are given the lengths of three non-parallel edges meeting at a vertex, call them x, y, and z, and the three angles between them which I will call a, b, and cs (for a reason that will be apparent later). The values don't matter, so make them up on your own.

I had my attention brought to this problem by an old journal article I was reading. The method caught my eye when I noticed that embedded in the formula was a trig looking version of Heron's (some say Hero's) formula for the area of a triangle. In the 1st century of the common era, Heron wrote a formula for the area of a triangle in his metrica. The formula, often appearing in High School texts, involves the semi perimeter of the triangle, s=$\frac{a+b+c}{2}$ where the lengths of the sides are a, b, and c. The area is given by $\sqrt{s(s-a)(s-b)(s-c)}$. The method was extended by Brahmagupta in the 7th century to give the area of any cyclic quadrilateral (one that is inscribed in a circle).

But for our case, it shows up as a relationship about the angles of the parallelepiped. For a parallelapiped with edges of length x, y, and z; which meet at mutual angles of a, b, and c, let s=$\frac{a+b+c}{2}$ as before, except now we are adding the angles. The volume is given by $2xyz\sqrt{sin(s)sin(s-a)sin(s-b)sin(s-c)}$. I kind of love when a "look-alike" formula pops up.

It reminds me of one more that is not well known to modern high school teachers. There is another variation of the law of sines for triangles in a plane $\frac{a}{sin(A)}=\frac{b}{Sin(B)}=\frac{c}{Sin(C)}$ (I really hate when that is written upside down).

It has a look-alike that works for spherical triangles; $\frac{sin(a)}{sin(A)}=\frac{sin(b)}{Sin(B)}=\frac{sin(c)}{Sin(C)}$. The spherical law of sines was first presented by Johann Muller, also known as Regiomontus,in his De Triangulis Omnimodis in 1464. This was the first book devoted wholly to trigonometry (a word not then invented). David E. Smith suggests that the theorem was Muller's invention.
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