Question Number 175986 by ajfour last updated on 10/Sep/22
Answered by a.lgnaoui last updated on 11/Sep/22
![∡AOB=∡FDC OA ∣∣DF CD ∣∣OB tan (θ)=((BF)/(BP))=(a/(BP)) a=BPtan (θ) ((BP)/(CD))=((BF)/(FC)) (a/(btan (θ)))=(a/(FC))=(a/(a+BC)) BC=DPsin (θ) DP=OA=R (a/(btan (θ)))=(a/(a+Rsin (θ))) (1/(btan (θ)))=(1/(a+Rsin (θ))) a+Rsin (θ)=btan (θ)⇒ btan (θ)−Rsin (θ)=a sin (θ)[(b/(cos( θ)))−R]=a ⇒ R=(b/(cos (θ)))−(a/(sin (θ))) a+b=(b/(cos (θ)))−(a/(sin (θ))) b((1/(cos (θ)))−1)−a((1/(sin (θ)))+1)=0 (b/a)=[((1+(1/(sin (θ))))/((1/(cos (θ)))−1))] (b/(cos (θ)))=(b/(b/(DF)))=DF (a/(sin (θ)))=(a/(a/(PF)))=PF donc R=DF−PF=DP](Q176036.png)
$$\measuredangle{AOB}=\measuredangle{FDC}\:\:\:\:{OA}\:\mid\mid{DF}\:\:\:\:{CD}\:\mid\mid{OB} \\ $$$$\mathrm{tan}\:\left(\theta\right)=\frac{\mathrm{BF}}{\mathrm{BP}}=\frac{\mathrm{a}}{\mathrm{BP}}\:\:\:\:\mathrm{a}=\mathrm{BPtan}\:\left(\theta\right) \\ $$$$\frac{\mathrm{BP}}{\mathrm{CD}}=\frac{\mathrm{BF}}{\mathrm{FC}}\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{btan}\:\left(\theta\right)}=\frac{{a}}{\mathrm{FC}}=\frac{{a}}{{a}+\mathrm{BC}} \\ $$$$\mathrm{BC}=\mathrm{DPsin}\:\left(\theta\right)\:\:\:\mathrm{DP}=\mathrm{OA}={R}\:\:\: \\ $$$$\:\frac{{a}}{{b}\mathrm{tan}\:\left(\theta\right)}=\frac{{a}}{{a}+{R}\mathrm{sin}\:\left(\theta\right)} \\ $$$$\frac{\mathrm{1}}{{b}\mathrm{tan}\:\left(\theta\right)}=\frac{\mathrm{1}}{{a}+{R}\mathrm{sin}\:\left(\theta\right)} \\ $$$${a}+{R}\mathrm{sin}\:\left(\theta\right)={b}\mathrm{tan}\:\left(\theta\right)\Rightarrow\:\:\:{b}\mathrm{tan}\:\left(\theta\right)−{R}\mathrm{sin}\:\left(\theta\right)={a} \\ $$$$\mathrm{sin}\:\left(\theta\right)\left[\frac{{b}}{\mathrm{cos}\left(\:\theta\right)}−{R}\right]={a}\:\:\:\:\:\Rightarrow\:\:\:\:\:\:{R}=\frac{{b}}{\mathrm{cos}\:\left(\theta\right)}−\frac{{a}}{\mathrm{sin}\:\left(\theta\right)} \\ $$$${a}+{b}=\frac{{b}}{\mathrm{cos}\:\left(\theta\right)}−\frac{{a}}{\mathrm{sin}\:\left(\theta\right)} \\ $$$${b}\left(\frac{\mathrm{1}}{\mathrm{cos}\:\left(\theta\right)}−\mathrm{1}\right)−{a}\left(\frac{\mathrm{1}}{\mathrm{sin}\:\left(\theta\right)}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{{b}}{{a}}=\left[\frac{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{sin}\:\left(\theta\right)}}{\frac{\mathrm{1}}{\mathrm{cos}\:\left(\theta\right)}−\mathrm{1}}\right] \\ $$$$\: \\ $$$$\frac{{b}}{\mathrm{cos}\:\left(\theta\right)}=\frac{{b}}{\frac{{b}}{\mathrm{DF}}}=\mathrm{DF}\:\:\:\:\:\frac{{a}}{\mathrm{sin}\:\left(\theta\right)}=\frac{{a}}{\frac{{a}}{\mathrm{PF}}}=\mathrm{PF} \\ $$$${donc}\:\:\:\:{R}=\mathrm{DF}−\mathrm{PF}=\mathrm{DP} \\ $$
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
