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 1010 = 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.