/sci/ - Science & Math

>>11562345
Let [math] S [/math] be the thing we are after. Let [math] a=\sin\theta [/math] Notice that if you scale triangle ABC down by a factor of BE/BC, you get S scaled down by the same amount. So [math] S-|CD|-|DE|=(|BE|/|BC|)S [/math]. Trigonometry gives [math] |CD|=ba [/math] and [math] |DE|=ba^2 [/math]. You can compute [math] |BE|=b\tan\theta-ba\cos\theta [/math] and [math] |BC|=b\tan\theta [/math]. Then you have [math] S-ba-ba^2=S\big[(b\tan\theta-ba\tan\cos\theta)/(b\tan\theta)\big]=Sa^2[/math]. Anyway, you can simplify: [eqn]
S=\frac{b(a+a^2)}{(1-a)^2}=\frac{ba}{1-a}=\frac{b\sin\theta}{1-\sin \theta} [/eqn] No need to write out an infinite sum!
>>11562468
Gross!

>>11562345
Let [math] S [/math] be the thing we are after. Let [math] a=\sin\theta [/math] Notice that if you scale triangle ABC down by a factor of BE/CE, you get S scaled down by the same amount. So [math] S-|CD|-|DE|=(|BE|/|CE|)S [/math]. Trigonometry gives [math] |CD|=ba [/math] and [math] |DE|=ba^2 [/math]. You can compute [math] |BE|=b\tan\theta-ba\cos\theta [/math] and [math] |BC|=b\tan\theta [/math]. Then you have [math] S-ba-ba^2=S\big[(b\tan\theta-ba\tan\cos\theta)/(b\tan\theta)\big]=Sa^2[/math]. Anyway, you can simplify: [eqn]
S=\frac{b(a+a^2)}{(1-a)^2}=\frac{ba}{1-a}=\frac{b\sin\theta}{1-\sin \theta} [/eqn] No need to write out an infinite sum!

>>11540315
Basically just consider the forces on point G. There are four forces: There is weight, whose direction and magnitude are known; then there are the reactions from A, B, C, and whose directions are all known (along the supports), you just need to find magnitude. Sum the four force vectors together, and you get a system of three equations for the three coordinates with three unknowns (two unknowns, if you consider symmetry).

>>11520796
Großwachsen!
>>11510441
The intersection of [math] f(x)=A\exp(x+A)+x [/math] with the x axis.
>>11507288
Excel or Octave (MATLAB). R if you are a chemical tranner.