OpenEuphoria

Re: trig question

• Posted by Matthew Lewis <MatthewL at KAPCOUSA.COM> Aug 31, 2000
• 799 views

Lewis Townsend wrote:

> This is a trig question I think. How do I find an angle
> of a right triangle if all I know is the length of the
> sides?
>=20
> later,
> Lewis Townsend
>=20
>   ?=B0
>   |\
>   | \sqrt((5*5)+(3*3))
> 5 |  \
>   |___\
> 90=B0 3  ?=B0

You need:

Law of Cosines: a^2 =3D b^2 + c^2 - 2bc cos A
Where a,b,c are the lengths of the sides, and A is the angle opposite =
a.

To solve for A:
A =3D arccos((a^2-b^2-c^2)/(-2bc))

Then B =3D 90-A
and, of course, C=3D90

So for your example:
a =3D 5, b =3D 3, c^2 =3D 34
A =3D arccos( (25-9-36)/(-2*sqrt(34)) )
a ~ arccos( 1.7 )
a ~ arccos( .7 )
(must be between -1 and 1)
a ~ .795rad
a ~ 45
(Some rounding here...)

Matt