**excerpts from**

**Benjamin Banneker's Trigonometry
Puzzle**

by Florence Fasanelli, Graham Jagger, Bea Lumpkin

**for complete article, see: http://convergence.mathdl.org/convergence/?pa=content&sa=viewDocument&nodeId=212**

Benjamin Banneker's Trigonometry Puzzle The mathematical puzzles of Benjamin Banneker (1731-1806) have been of interest since he first produced them in the last half of the 18th century and the early years of the 19th. By the age of 21, he was a hero in the territory of Maryland.1 Banneker began his life-changing studies of astronomy and mathematics about 1788, the year Maryland joined the Union, when he was lent some books by his friend the surveyor, George Ellicott. Three of these are known: Charles Leadbetter's Astronomy: Or the True System of the Planets Demonstrated (1727); James Ferguson's Astronomy Explained Upon Sir Isaac Newton's Principles, and Made Easy to Those Who Have Not Studied Mathematics(1761), and a book (in Latin) which Ellicott gave his bride as a wedding present.

In one of the puzzles, "Trigonometry," Banneker demonstrates his knowledge of logarithms as he presents his solution. The question arises as to what book of logarithmic tables could Banneker have been using. The trigonometry page is reproduced from The Maryland Historical Society, Baltimore, Maryland, where Banneker's Astronomical Journal 1798, is kept.

It is clear that Banneker is using the Law of Sines:

In a triangle, the ratios of the sine of an angle to the length of its opposite side are equal. Where Banneker writes in his proportion, "logarithm base 26," or "logarithm of the hypotenuse," he is anticipating the use of logarithms for the computation. Banneker understood, as his calculations correctly show, that the ratios involved are the sine of an angle to the side opposite, not to the log of the side. In what follows, all angles are expressed in degrees.

**I.** To find the
hypotenuse, Banneker used the Law of Sines: sin C/c = sin B/ b
"Sine complement of the angle at A," is sin 60, the
sine of the angle complementary to angle A. If x is the length
of the hypotenuse, then sin 60/ 26 = sin 90/ x and x = 26 sin
90/sin 60. . Taking logarithms, we have log x = log 26 + log sin
90 ??- log sin 60 and, substituting values from a suitable set
of tables, log x = 1.41497 +10 9.93753 = 1.47744. Notice
that, for reasons we shall see later, log sin 90 = Log 10^{10}
= 10. We now find x as the antilogarithm of 1.47744, which is
very close to 30. It is unlikely that Banneker would have had
access to tables of antilogarithms, a late eighteenth century
invention, but would simply have used his table of logarithms
in reverse.

**II.** To find the
remaining side, Banneker uses the Law of Sines again: sin C/c
= sin A/a. If x is the length of the side perpendicular to the
base then sin 60/ 26 = sin 30/ x, and x = 26 sin 30/sin 60 . Taking
logarithms we get log x = log 26 + log sin 30 - log sin 60 and,
again substituting values from tables, log x = 1.41497 + 9.69897
- 9.93753 = 1.7641. Again, x is the antilogarithm of 1.17643,
which is very close to 15.