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Question Number 175986 by ajfour last updated on 10/Sep/22

Answered by a.lgnaoui last updated on 11/Sep/22

$$\measuredangle AOB=\measuredangle FDCOA\mid \mid DFCD\mid \mid OB$$$$\tan \left(\theta \right)=\frac{\mathrm{BF}}{\mathrm{BP}}=\frac{\mathrm{a}}{\mathrm{BP}}\mathrm{a}=\mathrm{BPtan}\left(\theta \right)$$$$\frac{\mathrm{BP}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{FC}}\frac{\mathrm{a}}{\mathrm{btan}\left(\theta \right)}=\frac{a}{\mathrm{FC}}=\frac{a}{a+\mathrm{BC}}$$$$\mathrm{BC}=\mathrm{DPsin}\left(\theta \right)\mathrm{DP}=\mathrm{OA}=R$$$$\frac{a}{b\tan \left(\theta \right)}=\frac{a}{a+R\sin \left(\theta \right)}$$$$\frac{1}{b\tan \left(\theta \right)}=\frac{1}{a+R\sin \left(\theta \right)}$$$$a+R\sin \left(\theta \right)=b\tan \left(\theta \right)\Rightarrow b\tan \left(\theta \right)-R\sin \left(\theta \right)=a$$$$\sin \left(\theta \right)[\frac{b}{\cos \left(\theta \right)}-R]=a\Rightarrow R=\frac{b}{\cos \left(\theta \right)}-\frac{a}{\sin \left(\theta \right)}$$$$a+b=\frac{b}{\cos \left(\theta \right)}-\frac{a}{\sin \left(\theta \right)}$$$$b(\frac{1}{\cos \left(\theta \right)}-1)-a(\frac{1}{\sin \left(\theta \right)}+1)=0$$$$\frac{b}{a}=\left[\frac{1+\frac{1}{\sin \left(\theta \right)}}{\frac{1}{\cos \left(\theta \right)}-1}\right]$$$$$$$$\frac{b}{\cos \left(\theta \right)}=\frac{b}{\frac{b}{\mathrm{DF}}}=\mathrm{DF}\frac{a}{\sin \left(\theta \right)}=\frac{a}{\frac{a}{\mathrm{PF}}}=\mathrm{PF}$$$$doncR=\mathrm{DF}-\mathrm{PF}=\mathrm{DP}$$

Commented by a.lgnaoui last updated on 11/Sep/22

Commented by Tawa11 last updated on 15/Sep/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 12/Sep/22

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