Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

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## 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) ???