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Hi.
How do I solve sin(θ) = 0.38cos(θ) ???
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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
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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.
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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
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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.
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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
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Failure.
Boy let me tell you what:
I bet you didn't know it, but I'm a fiddle player too.
And if you'd care to take a dare, I'll make a bet with you.
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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...)
the problem starts when (using arcsin, arccos and arctan) you need to know the quadrant for angles larger than 90degrees (afaik)
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