EPT![[image loading]](http://i26.tinypic.com/eqeelz.jpg)
x=6.0
y=12.0
L1=12"
L2=10"
r=13.41
theta=63.43
theta1=
theta2=
Is this picture better?
The answers for theta1 is 109.4 and theta2 is 105.5 I just don't know where that came from.
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Raithed
China7078 Posts
x=6.0 y=12.0 L1=12" L2=10" r=13.41 theta=63.43 theta1= theta2= Is this picture better? The answers for theta1 is 109.4 and theta2 is 105.5 I just don't know where that came from. | ||
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ShOoTiNg_SpElLs
Korea (South)690 Posts
by cosine law, we get R^2 = L1^2 + L2^2 - 2*L1*L2 * cosx solving, we get x = 74.49~ since x and theta2 add up to 180, we get theta2 is 105.5 let the angle between L1 and R be y by cosine law, we get L2^2 = L1^2 + R^2 - 2 * L1 * R * cosy solving, we get y = 45.94~ since theta and y add up to theta1, we get theta1 is their sum i.e. 109.4 | ||
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Purind
Canada3562 Posts
Components is way easier than cosine law if you have more than 2 arms | ||
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ShOoTiNg_SpElLs
Korea (South)690 Posts
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Raithed
China7078 Posts
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ShOoTiNg_SpElLs
Korea (South)690 Posts
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Raithed
China7078 Posts
R^2 = L1^2 + L2^2 - 2*L1*L2 * cosx solving, we get x = 74.49~ cosx, it's not the same as cos(gamma)? Because I don't remember how to do this, I mean, which way do you look at the angle? | ||
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ShOoTiNg_SpElLs
Korea (South)690 Posts
R^2 = L1^2 + L2^2 - 2*L1*L2*cos(gamma) x is just a variable, as would gamma be btw when I wrote cosx = cos(x), if it's the (lack of) brackets confusing you | ||
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