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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 

  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 

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

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


and use that!  (grin)

-Doctor Rob,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
Associated Topics:
College Trigonometry
High School Trigonometry

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