Question Number 191192 by ajfour last updated on 20/Apr/23
Commented by ajfour last updated on 20/Apr/23
$${If}\:{lower}\:{circle}\:{has}\:{radius}\:{a}\:{while} \\ $$$${upper}\:{has}\:{radius}\:{b},\:{find}\:\theta. \\ $$
Answered by mr W last updated on 20/Apr/23
Commented by mr W last updated on 20/Apr/23
$${OA}=\frac{{a}}{\mathrm{tan}\:\theta} \\ $$$${AB}=\sqrt{\left({a}+{b}\right)^{\mathrm{2}} −\left({a}−{b}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{{ab}} \\ $$$${OB}=\frac{{a}}{\mathrm{tan}\:\theta}+\mathrm{2}\sqrt{{ab}}=\frac{{b}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$${let}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=\frac{\mathrm{2}{t}}{\mathrm{1}−{t}^{\mathrm{2}} } \\ $$$$\frac{{a}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}}+\mathrm{2}\sqrt{{ab}}=\frac{{b}}{{t}} \\ $$$${t}^{\mathrm{2}} −\mathrm{4}\sqrt{\frac{{b}}{{a}}}{t}+\frac{\mathrm{2}{b}}{{a}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{t}=\mathrm{2}\sqrt{\frac{{b}}{{a}}}−\sqrt{\frac{\mathrm{2}{b}}{{a}}+\mathrm{1}} \\ $$$$\Rightarrow\theta=\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \left(\mathrm{2}\sqrt{\frac{{b}}{{a}}}−\sqrt{\frac{\mathrm{2}{b}}{{a}}+\mathrm{1}}\right) \\ $$
Commented by ajfour last updated on 20/Apr/23
$${OB}=\frac{{a}}{\mathrm{tan}\:\theta}+\mathrm{2}\sqrt{{ab}}=\frac{{b}}{\mathrm{tan}\:\frac{\theta}{\mathrm{2}}} \\ $$$$\Rightarrow\:\:\mathrm{1}−{t}^{\mathrm{2}} +\mathrm{4}{t}\sqrt{\frac{{b}}{{a}}}=\mathrm{2}\left(\frac{{b}}{{a}}\right) \\ $$$$\Rightarrow\:{t}^{\mathrm{2}} −\mathrm{4}{t}\sqrt{\frac{{b}}{{a}}}+\frac{\mathrm{2}{b}}{{a}}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\mathrm{2}\sqrt{\frac{{b}}{{a}}}\pm\sqrt{\frac{\mathrm{2}{b}}{{a}}+\mathrm{1}} \\ $$$$\theta=\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{2}\sqrt{\frac{{b}}{{a}}}−\sqrt{\frac{\mathrm{2}{b}}{{a}}+\mathrm{1}}\right) \\ $$$${thanks}\:{sir}! \\ $$
Commented by mr W last updated on 20/Apr/23
$${yes},\:{you}\:{are}\:{right}\:{sir}. \\ $$