Question 753327
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I never actually saw such a thing, but there is no reason we can't derive such a formula.


Let the ratio of *[tex \LARGE a] to *[tex \LARGE b] be represented by *[tex \LARGE \frac{p}{q}], such that *[tex \LARGE \frac{a}{b}\ =\ \frac{p}{q}].


From this we can say *[tex \LARGE a\ =\ \frac{p}{q}b]


But we know that *[tex \LARGE a^2\ +\ b^2\ =\ c^2]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{p}{q}b\right)^2\ +\ b^2\ =\ c^2]


Solve for *[tex \LARGE b]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ \sqrt{\frac{q^2c^2}{p^2\,+\,q^2}}]


Then substitute *[tex \LARGE b] back into *[tex \LARGE a\ =\ \frac{p}{q}b] to calculate the value of *[tex \LARGE a]


In your example, presuming you mean that *[tex \LARGE a] is to *[tex \LARGE b] as 2.5 is to 1, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ b\ =\ \sqrt{\frac{289}{7.25}}\ \ \ ] so *[tex \LARGE \ \ a\ =\ 2.5\sqrt{\frac{289}{7.25}}]


And so on...


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
<font face="Math1" size="+2">Egw to Beta kai to Sigma</font>
My calculator said it, I believe it, that settles it
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