You are not logged in.
Post a reply
Topic review (newest first)
- Nitrofu
- 2005-07-28 19:46:40
the problem starts when (using arcsin, arccos and arctan) you need to know the quadrant for angles larger than 90degrees (afaik)
- Nitrofu
- 2005-07-28 19:43:10
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...)
- Zach
- 2005-07-24 22:23:28
Failure.
- MathsIsFun
- 2005-07-24 22:00:01
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!
- mathsyperson
- 2005-07-24 21:37:30
You can time how long it takes people to post?
That's extremely cool/pointless.
- MathsIsFun
- 2005-07-24 21:19:36
SNAP 
(Looking at the times, I must be slower at finishing my posts)
- mathsyperson
- 2005-07-24 21:17:40
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
- MathsIsFun
- 2005-07-24 21:14:32
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°
- GurraTedden
- 2005-07-24 21:10:22
Hi.
How do I solve sin(θ) = 0.38cos(θ) ???