Computing Angles of a Right Triangle
Date: 09/04/97 at 15:17:39 From: Jayme Silva Subject: Computing Angles of a Right Triangle From your archives: "How do you compute the other two angles of a right triangle? I know the other two lengths, but how do I get the other two angles?" The answer, from Dr. Tom, is "push the arctan button on your calculator"! Come on, I have neither a scientific calculator, nor a slide rule, nor a book of tables. Instead, I have my head and a pencil and paper. Is there a way to do this manually? If a table is required, where can I find it? Thanks.
Date: 09/04/97 at 16:23:29 From: Doctor Rob Subject: Re: Computing Angles of a Right Triangle There are tables in every book on trigonometry ever written. There are also tables in the Handbook of Chemistry and Physics (old) or the Chemical Rubber Company Tables (newer). If you insist on not doing it with tables, you can compute the arctangent function using Maclaurin series as follows: arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + x^9/9 - ... (-1 < x < 1). If you truncate this series, the error will be less than the first omitted term. The answer will appear in radians, so you will need to multiply by 180/Pi to convert to degrees. Admittedly this involves quite a bit of calculation using pencil and paper, but it is feasible. The smaller x is, the faster this series converges. If x is very close to 1, so that the series converges too slowly, you can use the formula arctan(x) = 2*arctan[(sqrt[1+x^2]-1)/x], and the argument of the arctangent on the right is farther from 1 so that the Maclaurin series will converge faster. This process can be iterated. For very small values of x, arctan(x) = x is a good approximation. For example, if you find that you need the arctangent of .32000, you can compute x = .32000 x = .32000 x^2 = .10240 x^3 = .03277 -x^3/3 = -.01092 x^5 = .00336 x^5/5 = .00067 x^7 = .00034 -x^7/7 = -.00005 x^9 = .00004 x^9/9 = .000004 sum = .30970, accurate to 5 significant figures, arctan(x) = .30970 radians, = .30970*180/Pi = 17.745 degrees. If x is extremely close to 1, here is a method which is actually better: When x is near the tangent of an angle A you know, such as x near 1 and A = Pi/4, then you can do this: y = arctan(x), x = tan(y), tan(y-A) = [tan(y)-tan(A)]/[1+tan(y)*tan(A)], y = A + arctan([x-tan(A)]/[1+x*tan(A)]), arctan(x) = A + arctan([x-tan(A)]/[1+x*tan(A)]). The fraction inside the arctan on the right is quite small, and the convergence of the Maclaurin series is accordingly very fast. In the case where A = Pi/4, one is faced with finding the arctangent of (x-1)/(x+1), which is very small. If your x > 1, then 1/x < 1, and you need to use the relation arctan(x) = Pi/2 - arctan(1/x). Or you can access the Online Scientific Calculator at http://huizen.dds.nl/~chronix/calc.htm and use that! (grin) -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2013 The Math Forum