Question Number 166465 by ajfour last updated on 20/Feb/22
Commented by ajfour last updated on 20/Feb/22
$${Find}\:\frac{{a}}{{b}}\:\:{in}\:{terms}\:{of}\:\alpha\:{and}\:\beta. \\ $$
Answered by mr W last updated on 20/Feb/22
Commented by mr W last updated on 20/Feb/22
$${R}={radius}\:{of}\:{circle} \\ $$$${CD}=\mathrm{2}{R} \\ $$$${AC}=\mathrm{2}{R}\:\mathrm{cos}\:\alpha \\ $$$${AC}=\frac{{a}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\frac{{a}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)} \\ $$$$\mathrm{2}{R}\:\mathrm{cos}\:\alpha={a}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)}\right) \\ $$$${similarly} \\ $$$$\mathrm{2}{R}\:\mathrm{cos}\:\beta={b}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}}\right)}\right) \\ $$$$\Rightarrow\frac{{a}}{{b}}=\frac{\mathrm{cos}\:\alpha\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}+\frac{\mathrm{1}+\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\right)}{\mathrm{cos}\:\beta\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\frac{\mathrm{1}+\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}\right)} \\ $$
Commented by ajfour last updated on 20/Feb/22
$${perfect}!\:{thank}\:{you}\:{sir}. \\ $$