OpenEuphoria

Re: trig question

• Posted by Derek Parnell <dparnell at BIGPOND.NET.AU> Sep 01, 2000
• 790 views

----- Original Message -----
From: "Lewis Townsend" <keroltarr at HOTMAIL.COM>
To: <EUPHORIA at LISTSERV.MUOHIO.EDU>
Sent: Friday, September 01, 2000 5:53 AM
Subject: trig question


> Hello all,
>
> 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?
>
> later,
> Lewis Townsend
>
>   ?=B0
>   |\
>   | \sqrt((5*5)+(3*3))
> 5 |  \
>   |___\
> 90=B0 3  ?=B0

I'm going back a few years to my high school nightmares. My maths teacher
drummed this acronym into our heads to help us remember "SOH-CAH-TOA"

SOH =3D=3D> Sin(x)=3D(Opposite / Hypotenuse)

CAH =3D=3D> Cos(x)=3D(Adjacent / Hypotenuse)

TOA =3D=3D> Tan(x)=3D(Opposite / Adjacent)

Thus...

   a=B0
   |\
   | \sqrt((5*5)+(3*3))
 5 |  \
   |___\
 90=B0 3  b=B0

Sin(a) =3D (3 / sqrt((5*5)+(3*3)))
Sin(a) =3D 0.5145
a =3D ArcSin(0.5145)
a =3D 0.5404 radians
a =3D (0.5404 * 180) / PI degrees
a =3D 30.9640 degrees

b =3D 90 - a
b =3D 59.0360 degrees