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

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**GurraTedden****Member**- Registered: 2005-07-20
- Posts: 19

Hi.

How do I solve sin(θ) = 0.38cos(θ) ???

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

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

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

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

Offline

**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

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.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,529

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

Offline

**Zach****Member**- Registered: 2005-03-23
- Posts: 2,075

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.

Offline

**Nitrofu****Guest**

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...)

**Nitrofu****Guest**

the problem starts when (using arcsin, arccos and arctan) you need to know the quadrant for angles larger than 90degrees (afaik)

Pages: **1**