Question Number 125508 by ajfour last updated on 11/Dec/20
Commented by ajfour last updated on 11/Dec/20
$${Find}\:{R}. \\ $$
Answered by ajfour last updated on 11/Dec/20
Commented by ajfour last updated on 11/Dec/20
$${let}\:\angle{C}=\mathrm{2}\theta \\ $$$${AE}={AF}+{EF} \\ $$$${R}\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)=\mathrm{tan}\:\left(\frac{\pi}{\mathrm{4}}+\theta\right)+\mathrm{2}\sqrt{{R}} \\ $$$$\:\:\:…..\left({I}\right) \\ $$$${CE}={ED}+{DC} \\ $$$${R}\mathrm{cot}\:\mathrm{2}\theta=\mathrm{2}\sqrt{{R}}+\mathrm{cot}\:\theta\:\:\:\:\:\:….\left({II}\right) \\ $$$${let}\:\:\mathrm{tan}\:\theta={m} \\ $$$${from}\:\left({I}\right): \\ $$$${R}=\mathrm{1}+\mathrm{2}\sqrt{{R}}\left(\frac{\mathrm{1}−{m}}{\mathrm{1}+{m}}\right) \\ $$$$\Rightarrow\:\:\frac{{R}−\mathrm{1}}{\mathrm{2}\sqrt{{R}}}=\frac{\mathrm{1}−{m}}{\mathrm{1}+{m}} \\ $$$${m}=\frac{\mathrm{2}\sqrt{{R}}+\mathrm{1}−{R}}{\mathrm{2}\sqrt{{R}}+{R}−\mathrm{1}} \\ $$$${and}\:{from}\:\left({II}\right): \\ $$$${R}=\mathrm{2}\sqrt{{R}}\left(\frac{\mathrm{2}{m}}{\mathrm{1}−{m}^{\mathrm{2}} }\right)+\frac{\mathrm{2}}{\mathrm{1}−{m}^{\mathrm{2}} } \\ $$$$\Rightarrow \\ $$$$\:\:{R}\left\{\mathrm{1}−\left(\frac{\mathrm{2}\sqrt{{R}}+\mathrm{1}−{R}}{\mathrm{2}\sqrt{{R}}+{R}−\mathrm{1}}\right)^{\mathrm{2}} \right\}=\mathrm{4}\sqrt{{R}}\left(\frac{\mathrm{2}\sqrt{{R}}+\mathrm{1}−{R}}{\mathrm{2}\sqrt{{R}}+{R}−\mathrm{1}}\right)+\mathrm{2} \\ $$$$\Rightarrow \\ $$$$\mathrm{4}{R}\sqrt{{R}}\left({R}−\mathrm{1}\right)=\mathrm{2}\sqrt{{R}}\left\{\mathrm{4}{R}−\left({R}−\mathrm{1}\right)^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{4}{R}+\left({R}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\sqrt{{R}}\left({R}−\mathrm{1}\right) \\ $$$${say}\:{R}={x} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \sqrt{{x}}−\mathrm{4}{x}\sqrt{{x}}=\mathrm{16}{x}\sqrt{{x}}−\mathrm{6}\sqrt{{x}}+\mathrm{4}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} \sqrt{{x}} \\ $$$$\Rightarrow\:\:\mathrm{6}\sqrt{{x}}\left({x}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{20}{x}\sqrt{{x}}+\left({x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\Rightarrow\:\:{x}\approx\:\mathrm{3}.\mathrm{56918} \\ $$$$\left({Sir},\:{i}\:{think}\:{just}\:{this}\:{answer}\right. \\ $$$$\left.\:\:\:{is}\:{valid}\right) \\ $$
Commented by mr W last updated on 11/Dec/20
$${yes},\:{you}\:{are}\:{right}\:{sir}! \\ $$
Answered by mr W last updated on 11/Dec/20
Commented by mr W last updated on 11/Dec/20
![sin α=((R−1)/(R+1)) sin β=(1/(R+1)) γ=α−β=(π/2)−2α ⇒3α=(π/2)+β ⇒sin (3α)=sin ((π/2)+β)=cos β ⇒sin α (3−4 sin^2 α)=cos β ⇒(((R−1)/(R+1)))[3−4×(((R−1)/(R+1)))^2 ]=((√((R+1)^2 −1))/(R+1)) ⇒(R−1)[3−4×(((R−1)/(R+1)))^2 ]=(√(R(R+2))) ⇒R≈2.3981 or 3.5692](https://www.tinkutara.com/question/Q125518.png)
$$\mathrm{sin}\:\alpha=\frac{{R}−\mathrm{1}}{{R}+\mathrm{1}} \\ $$$$\mathrm{sin}\:\beta=\frac{\mathrm{1}}{{R}+\mathrm{1}} \\ $$$$\gamma=\alpha−\beta=\frac{\pi}{\mathrm{2}}−\mathrm{2}\alpha \\ $$$$\Rightarrow\mathrm{3}\alpha=\frac{\pi}{\mathrm{2}}+\beta \\ $$$$\Rightarrow\mathrm{sin}\:\left(\mathrm{3}\alpha\right)=\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}+\beta\right)=\mathrm{cos}\:\beta \\ $$$$\Rightarrow\mathrm{sin}\:\alpha\:\left(\mathrm{3}−\mathrm{4}\:\mathrm{sin}^{\mathrm{2}} \:\alpha\right)=\mathrm{cos}\:\beta \\ $$$$\Rightarrow\left(\frac{{R}−\mathrm{1}}{{R}+\mathrm{1}}\right)\left[\mathrm{3}−\mathrm{4}×\left(\frac{{R}−\mathrm{1}}{{R}+\mathrm{1}}\right)^{\mathrm{2}} \right]=\frac{\sqrt{\left({R}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}}{{R}+\mathrm{1}} \\ $$$$\Rightarrow\left({R}−\mathrm{1}\right)\left[\mathrm{3}−\mathrm{4}×\left(\frac{{R}−\mathrm{1}}{{R}+\mathrm{1}}\right)^{\mathrm{2}} \right]=\sqrt{{R}\left({R}+\mathrm{2}\right)} \\ $$$$\Rightarrow{R}\approx\mathrm{2}.\mathrm{3981}\:{or}\:\mathrm{3}.\mathrm{5692} \\ $$
Commented by ajfour last updated on 11/Dec/20
$${Thanks}\:{Sir},\:{nice}\:{way}!\: \\ $$
Answered by islom last updated on 11/Feb/21
