x = 6.0
y = 12.0
L1 = 12"
L2 = 10"
R = sqrtof(10^2+12^2) = 15.62
theta = tan^-1(y/x) = tan^-1(12/6) = 63.43º
How do I find theta1, and theta2?
Sorry for the yellow:
![[image loading]](http://i25.tinypic.com/m7scz.jpg)





Blogs > Raithed |
Raithed
China7078 Posts
x = 6.0 y = 12.0 L1 = 12" L2 = 10" R = sqrtof(10^2+12^2) = 15.62 theta = tan^-1(y/x) = tan^-1(12/6) = 63.43º How do I find theta1, and theta2? Sorry for the yellow: ![]() ![]() ![]() ![]() ![]() ![]() | ||
hellhawk123
United States84 Posts
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Raithed
China7078 Posts
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Lemonwalrus
United States5465 Posts
Edit: oh, nvmd, I think. Edit2: oh, nvmd the nvmd. | ||
fight_or_flight
United States3988 Posts
On January 31 2008 11:31 Raithed wrote: No, R = sqrtof(10^2+12^2) = 15.62. Not x, y, but a, b. I don't remember those laws. :[ how is R equal to that? you can only use pythagorean theorem for a right-triangle. I'm kind of confused on the question, you should word it better. edit: lol at the theta | ||
Raithed
China7078 Posts
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hellhawk123
United States84 Posts
also law of cosines is given triangle w/ lengths a b c and respective angles alpha beta gamma, c^2=a^2+b^2-2abcos(gamma) (pythag theorem is when gamma=pi/2) | ||
fight_or_flight
United States3988 Posts
On January 31 2008 11:36 hellhawk123 wrote: sorry but how are you getting R? I assumed x=x component of R and y=y component of R? also law of cosines is given triangle w/ lengths a b c and respective angles alpha beta gamma, c^2=a^2+b^2-2abcos(gamma) Correct. If R is given by the (x,y) components, then it should be easy to plug in a, b, c and solve for gamma. | ||
Raithed
China7078 Posts
So basically it is: 13.42^2 = 10^2 + 12^2 - 2(10)(12)cos(gamma) 180 = 100+144 - 240cos(gamma) 180 = 244 - 240cos(gamma) Oh fuck I fail at Algebra, I know the gamma has to be alone, so move the (244 - 240cos) to the other side? | ||
hellhawk123
United States84 Posts
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Raithed
China7078 Posts
Thanks hellhawk, so that would be theta1? | ||
hellhawk123
United States84 Posts
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Raithed
China7078 Posts
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hellhawk123
United States84 Posts
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Kwidowmaker
Canada978 Posts
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Raithed
China7078 Posts
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Kwidowmaker
Canada978 Posts
I'm of the understanding that L1 is only the black and not the yellow line. | ||
Raithed
China7078 Posts
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BluzMan
Russian Federation4235 Posts
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Slayer91
Ireland23335 Posts
theta1 is theta minus the angel opposite l2. Finding those angles is a matter of plugging in the cosine rule, right? | ||
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