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Question Number 166137 by ajfour last updated on 13/Feb/22

Answered by mr W last updated on 14/Feb/22

Commented by mr W last updated on 14/Feb/22

$$\frac{\sin (\theta +\varphi )}{\sin \theta }=\frac{r+\frac{r}{\sin \varphi }}{2r}=\frac{1}{2}(1+\frac{1}{\sin \varphi })$$$$\frac{\sin \theta \cos \varphi +\cos \theta \sin \varphi }{\sin \theta }=\frac{1}{2}(1+\frac{1}{\sin \varphi })$$$$\cos \varphi +\frac{\sin \varphi }{\tan \theta }=\frac{1}{2}(1+\frac{1}{\sin \varphi })$$$$\frac{\sin \varphi }{\tan \theta }=\frac{1+\sin \varphi -\sin 2\varphi }{2\sin \varphi }$$$$\Rightarrow \tan \theta =\frac{1-\cos 2\varphi }{\sin \varphi +1-\sin 2\varphi }$$$${\left(\tan \theta \right)}_{max}\approx 1.5261$$$$\Rightarrow {\theta }_{max}\approx 56.7647°$$

Answered by MJS_new last updated on 14/Feb/22

$$\mathrm{I}\mathrm{tried}\mathrm{with}r=1\mathrm{the}\mathrm{center}\mathrm{of}\mathrm{the}\mathrm{circle}\left(\begin{array}{c}p\\ 1\end{array}\right)$$$$\mathrm{and}\Rightarrow △ABC$$$$A=\left(\begin{array}{c}0\\ 0\end{array}\right)B=\left(\begin{array}{c}2\\ 0\end{array}\right)C=\left(\begin{array}{c}p+\frac{p}{\sqrt{p^2+1}}\\ 1+\frac{1}{\sqrt{p^2+1}}\end{array}\right)$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{then}\mathrm{possible}\mathrm{to}\mathrm{get}\mathrm{an}\mathrm{exact}\mathrm{value}\mathrm{for}p$$$$\mathrm{with}\theta \to \max :$$$$p=\frac{1}{2}+\frac{\sqrt{5}}{4}-\frac{\sqrt{-7+4\sqrt{5}}}{4}\approx .710424116687$$$$\mathrm{and}\Rightarrow \theta \approx 56.7648°$$$$\mathrm{which}\mathrm{of}\mathrm{course}\mathrm{is}\mathrm{the}\mathrm{same}\mathrm{result}\mathrm{as}\mathrm{mrW}\mathrm{got}$$

Commented by mr W last updated on 15/Feb/22

$$greatsir!$$$$somehowihaveexpectedthatyou$$$$maygettheexactsolution.thanks!$$

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