1. [E - 18 to 19 Nov 1997] Triangles
- Posted by Jeff Zeitlin <jeff.zeitlin at MAIL.EXECNET.COM>
Nov 21, 1997
-
Last edited Nov 22, 1997
On Thu, 20 Nov 1997 00:02:07 -0500, Tuxedo Mask
<nitrogen at XTRA.CO.NZ> wrote:
>Hello, who knows the fastest way to find an angle of a right angled
>triangle, given the length of two of its sides. It doesn't need to be
>too acurate, so perhaps it can be made fast by using integer numbers for
>the results.
>eg:
> /|
> / |
> / |a
> / _|
>/______|_|
>A b C
>Lengths a and b are known, and angle C is 90 degrees.
>I need to find angle A.
The ratio a/b is equal to the tangent of A. Therefore, A is
equal to arctan(a/b).
If you know any two of the sides of a right triangle, you can
determine the length of the third side, and both of the "unknown"
angles:
B
/|
c / |
/ |a
/ _|
/______|_|
A b C
c*c =3D (a*a) + (b*b)
sin A =3D a/c
cos A =3D b/c
tan A =3D a/b
B =3D 90 - A
=46or all right triangles, sin A =3D cos (90 - A)
--
Jeff Zeitlin
jeff.zeitlin at mail.execnet.com
