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#2 2005-07-24 21:14:32
- MathsIsFun
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Re: Angle equation
I remember my trig using "SOHCAHTOA" (see http://www.mathsisfun.com/sine-cosine-tangent.html)
Therefore tan θ = sin θ / cos θ
So: sin(θ)/cos(θ) = 0.38 = tan θ
And: θ = tan-¹ 0.38 = 20.8°
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#3 2005-07-24 21:17:40
- mathsyperson
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Re: Angle equation
Divide both sides by cosθ: sinθ/cosθ=0.38
Simplify left-hand side: tanθ=0.38
Take arctans: θ=tan-¹(0.38)=20.8° to 1 decimal place
Why did the vector cross the road?
It wanted to be normal.
#4 2005-07-24 21:19:36
- MathsIsFun
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Re: Angle equation
SNAP 
(Looking at the times, I must be slower at finishing my posts)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#5 2005-07-24 21:37:30
- mathsyperson
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Re: Angle equation
You can time how long it takes people to post?
That's extremely cool/pointless.
Why did the vector cross the road?
It wanted to be normal.
#6 2005-07-24 22:00:01
- MathsIsFun
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Re: Angle equation
No, not really, though that is an interesting idea. It just seemed to me that there were 3 minutes difference between our posts, but I started first, but . . . oh, I don't know!
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman
#8 2005-07-28 19:43:10
- Nitrofu
- Guest
Re: Angle equation
the 3d .obj reader i have at http://www15.brinkster.com/nitrofurano/sdlbasic/ uses this
(i needed to rotate the viewer over the landscape, and for this i need to know the angle)
(this is simple to do as code, i don't know math notations as well...)
#9 2005-07-28 19:46:40
- Nitrofu
- Guest
Re: Angle equation
the problem starts when (using arcsin, arccos and arctan) you need to know the quadrant for angles larger than 90degrees (afaik)