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Question Number 166465 by ajfour last updated on 20/Feb/22

Commented by ajfour last updated on 20/Feb/22

Find (a/b)  in terms of α and β.

$${Find}\:\frac{{a}}{{b}}\:\:{in}\:{terms}\:{of}\:\alpha\:{and}\:\beta. \\ $$

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=2R  AC=2R cos α  AC=(a/(tan (α/2)))+(a/(tan ((π/4)−(β/2))))  2R cos α=a((1/(tan (α/2)))+(1/(tan ((π/4)−(β/2)))))  similarly  2R cos β=b((1/(tan (β/2)))+(1/(tan ((π/4)−(α/2)))))  ⇒(a/b)=((cos α((1/(tan (β/2)))+((1+tan (α/2))/(1−tan (α/2)))))/(cos β((1/(tan (α/2)))+((1+tan (β/2))/(1−tan (β/2))))))

$${R}={radius}\:{of}\:{circle} \\ $$$${CD}=\mathrm{2}{R} \\ $$$${AC}=\mathrm{2}{R}\:\mathrm{cos}\:\alpha \\ $$$${AC}=\frac{{a}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\frac{{a}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)} \\ $$$$\mathrm{2}{R}\:\mathrm{cos}\:\alpha={a}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\beta}{\mathrm{2}}\right)}\right) \\ $$$${similarly} \\ $$$$\mathrm{2}{R}\:\mathrm{cos}\:\beta={b}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}−\frac{\alpha}{\mathrm{2}}\right)}\right) \\ $$$$\Rightarrow\frac{{a}}{{b}}=\frac{\mathrm{cos}\:\alpha\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}+\frac{\mathrm{1}+\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}\right)}{\mathrm{cos}\:\beta\left(\frac{\mathrm{1}}{\mathrm{tan}\:\frac{\alpha}{\mathrm{2}}}+\frac{\mathrm{1}+\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}{\mathrm{1}−\mathrm{tan}\:\frac{\beta}{\mathrm{2}}}\right)} \\ $$

Commented by ajfour last updated on 20/Feb/22

perfect! thank you sir.

$${perfect}!\:{thank}\:{you}\:{sir}. \\ $$

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