Question 305563
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So far, so good.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{1}{3}\pi r^2h\ =\ 500]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r^2h\ =\ \frac{1500}{\pi}\ \approx\ 477.46]


but there are an infinite number of combinations of different values of *[tex \Large r] and *[tex \Large h] such that *[tex \Large r^2h\ \approx\ 477.46].


Then only thing you can do now is to fix one of the values and calculate the other.


For example, if you choose 10 for the radius, the height would come work out very close to 4.775


Or 5 for the radius gives you *[tex \Large h\ =\ \frac{60}{\pi}] -- not quite 20.


Or a height of 15 would give you a radius of *[tex \Large r\ =\ \frac{10}{\sqrt{\pi}}] -- A little over 5 and a half.


Now if you want a specific angle between the cone generator and the axis, that is another story altogether.  By the way, it is Theta, not "Theda"


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \theta\ =\ \tan^{-1}(\frac{r}{h})]


Which is to say that


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ h\tan(\theta)]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ r\cot(\theta)]


Which, combined with the earlier derivations leads us to:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\ =\ \sqrt[3]{\frac{1500}{\pi\cot\theta}}]


or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ h\ =\ \sqrt[3]{\frac{1500}{\pi\tan^2\theta}}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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