Question Number 223449 by ajfour last updated on 25/Jul/25
Answered by ajfour last updated on 26/Jul/25
$$\:\:\:\sqrt{\mathrm{1}−\theta^{\mathrm{2}} }=\frac{\mathrm{sin}\:\left(\phi−\theta\right)}{\mathrm{sin}\:\phi} \\ $$$${Wow}!\:\:\:\:\:\:{arc}\:{CD}=\:{line}\:{AB} \\ $$$$ \\ $$$$\bullet\:{Find}\:{CT}/{AO}={t}/{R}\:\:{in}\:{terms}\:{of}\:\theta. \\ $$$$\bullet\:{Find}\:\theta\:{whenAO}\:{is}\:{parallel}\:{to}\:{CT}. \\ $$
Answered by mr W last updated on 26/Jul/25
![((OT)/(OC))=((sin (φ−θ))/(sin φ)) ⇒OT=((R sin (φ−θ))/(sin φ)) TA^2 =(R+OT)(R−OT)=R^2 [1−((sin^2 (φ−θ))/(sin^2 φ))] KC=Rθ=TA R^2 θ^2 =R^2 [1−((sin^2 (φ−θ))/(sin^2 φ))] ((sin (φ−θ))/(sin φ))=(√(1−θ^2 )) tan φ=((sin θ)/(cos θ−(√(1−θ^2 )))) ⇒φ=tan^(−1) ((sin θ)/(cos θ−(√(1−θ^2 ))))](https://www.tinkutara.com/question/Q223457.png)
$$\frac{{OT}}{{OC}}=\frac{\mathrm{sin}\:\left(\phi−\theta\right)}{\mathrm{sin}\:\phi}\:\Rightarrow{OT}=\frac{{R}\:\mathrm{sin}\:\left(\phi−\theta\right)}{\mathrm{sin}\:\phi} \\ $$$${TA}^{\mathrm{2}} =\left({R}+{OT}\right)\left({R}−{OT}\right)={R}^{\mathrm{2}} \left[\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\left(\phi−\theta\right)}{\mathrm{sin}^{\mathrm{2}} \:\phi}\right] \\ $$$${KC}={R}\theta={TA} \\ $$$${R}^{\mathrm{2}} \theta^{\mathrm{2}} ={R}^{\mathrm{2}} \left[\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\left(\phi−\theta\right)}{\mathrm{sin}^{\mathrm{2}} \:\phi}\right] \\ $$$$\frac{\mathrm{sin}\:\left(\phi−\theta\right)}{\mathrm{sin}\:\phi}=\sqrt{\mathrm{1}−\theta^{\mathrm{2}} } \\ $$$$\mathrm{tan}\:\phi=\frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta−\sqrt{\mathrm{1}−\theta^{\mathrm{2}} }} \\ $$$$\Rightarrow\phi=\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{sin}\:\theta}{\mathrm{cos}\:\theta−\sqrt{\mathrm{1}−\theta^{\mathrm{2}} }} \\ $$