Question 746890
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Let *[tex \LARGE \beta] refer to angle APC.  Let *[tex \LARGE \alpha] refer to angle APB.  Our task is to maximize *[tex \LARGE \alpha\ -\ \beta]


*[tex \LARGE \tan\beta] is indeed *[tex \LARGE \frac{2}{1\ -\ t}], but you have a sign error or *[tex \LARGE \tan\alpha].  Should be *[tex \LARGE \tan\alpha\ =\ \frac{-1}{1\ +\ t}].


Using those values, and the fact that *[tex \LARGE \tan(\alpha\ -\ \beta)\ =\ \frac{\tan\alpha\ -\ \tan\beta}{1\ +\ \tan\alpha\,\tan\beta}], you should be able to derive the fact that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \tan(\alpha\ -\ \beta)\ =\ \frac{t\ +\ 3}{t^2\ +\ 1}]


Use the quotient rule to find the derivative,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d\left(\tan(\alpha\,-\,\beta)\right)}{dt}\ =\ \frac{-t^2\ -\ 6t\ +\ 1}{\left(t^2\ +\ 1\right)^2}]


Set the numerator of the derivative equal to zero and solve the quadratic for the positive value for *[tex \LARGE t] that yields a maximum tangent, and therefore a maximum angle.


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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