1. 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
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2. Re: trig question
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
3. Re: trig question
----- 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
4. Re: trig question
Thanks Matthew,
>You need:
>
>Law of Cosines: a^2 = 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 = arccos((a^2-b^2-c^2)/(-2bc))
>
>Then B = 90-A
>and, of course, C=90
later,
Lewis Townsend
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5. Re: trig question
Thanks for the info Derek... and anyone
else that responds to my question. I think
I can figure it out now.
Thanks again,
Lewis Townsend
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6. Re: trig question
On Fri, 1 Sep 2000 00:00:36 -0400, Lewis Townsend <keroltarr at HOTMAIL.COM>
wrote:
>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
Yes, it's trig. The 'magic word' for this is SOHCAHTOA. It refers to th=
e
method of calculating the three most common trigonometric functions. The
three sides of the triangle are labelled 'O' for 'Opposite', 'A' for
'Adjacent' and 'H' for 'Hypotenuse' The 'magic word' describes the
division that is used to compute the function, i.e., Sine is Opposite ove=
r
Hypotenuse' (for the top angle in your diagram, 3/sqrt(...)), Cosine is
Adjacent over Hypotenuse' (5/sqrt(...)), and Tangent is Opposite over
Adjacent (3/5). Then, look up the angle in a table for the appropriate
function. (Or use the arcsin, arccos, arctan functions on your calculator=
.)
A little bit of work might allow you to derive the 'direct' calculations;
IIRC, there is some stuff in zeitlin.zip in the Euphoria archives that ha=
s
(most of) the appropriate functions in it.
--
Jeff Zeitlin
jzeitlin at cyburban.com
