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Question-193511




Question Number 193511 by Lekhraj last updated on 15/Jun/23
Answered by a.lgnaoui last updated on 17/Jun/23
∡OAE=α   OA=OB=OF=R  ⇒∡OAF=∡OFA=α  ⇒cos α=((AF)/(2R))=((11)/(2R)) ⇒     R=((11)/(2cos 𝛂)) (1)  Surface triangle (AOE) est:             S=((OA×AEsin 𝛂)/2)=((11×4sin 𝛂)/(cos 𝛂))               S=44tan𝛂   (2)         avec { ((AE=16)),((OA=R=)) :}   •  triangle  ( OBF)   BF^2 =2R^2 (1−cos ϕ)   (ϕ=(π/2) −(π−2α)=2α−(π/2)   )  ⇒  BF^2 =2R^2 (1−sin 2𝛂)              (3)  •triangle (ABE)  AE⊥BE    (BE=r)  AB^2 =AE^2 +r^2 ⇒     2R^2 =16^2 +r^2   •BEF   BF^2 =BE^2 +EF^2 =r^2 +25                         =2R^2 −231                (4)     2R^2 (1−sin 2𝛂)=2R^2 −231     2R^2 sin 2𝛂−231=0       2(((11^2 )/(4cos^2 𝛂)))sin 2𝛂=231  (sin 2α=((2t)/(1+t^2 )),  (1/(cos^2 𝛂))=1+t^2 ;  t=tan α)        121t=231        t=((231)/(121))       (5)  dans l expression(2)         S=44tan 𝛂=((924)/(11))               Surface Triangle (AOE)  =84
$$\measuredangle\mathrm{OAE}=\alpha\:\:\:\mathrm{OA}=\mathrm{OB}=\mathrm{OF}=\mathrm{R} \\ $$$$\Rightarrow\measuredangle\mathrm{OAF}=\measuredangle\mathrm{OFA}=\alpha \\ $$$$\Rightarrow\mathrm{cos}\:\alpha=\frac{\mathrm{AF}}{\mathrm{2R}}=\frac{\mathrm{11}}{\mathrm{2R}}\:\Rightarrow\:\:\:\:\:\boldsymbol{\mathrm{R}}=\frac{\mathrm{11}}{\mathrm{2cos}\:\boldsymbol{\alpha}}\:\left(\mathrm{1}\right) \\ $$$$\mathrm{Surface}\:\mathrm{triangle}\:\left(\mathrm{AOE}\right)\:\mathrm{est}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}=\frac{\boldsymbol{\mathrm{OA}}×\boldsymbol{\mathrm{AE}}\mathrm{sin}\:\boldsymbol{\alpha}}{\mathrm{2}}=\frac{\mathrm{11}×\mathrm{4sin}\:\boldsymbol{\alpha}}{\mathrm{cos}\:\boldsymbol{\alpha}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}=\mathrm{44tan}\boldsymbol{\alpha}\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\mathrm{avec\begin{cases}{\boldsymbol{\mathrm{AE}}=\mathrm{16}}\\{\boldsymbol{\mathrm{OA}}=\boldsymbol{\mathrm{R}}=}\end{cases}} \\ $$$$\:\bullet\:\:\mathrm{triangle}\:\:\left(\:\mathrm{OBF}\right)\:\:\:\mathrm{BF}^{\mathrm{2}} =\mathrm{2R}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{cos}\:\varphi\right) \\ $$$$\:\left(\varphi=\frac{\pi}{\mathrm{2}}\:−\left(\pi−\mathrm{2}\alpha\right)=\mathrm{2}\alpha−\frac{\pi}{\mathrm{2}}\:\:\:\right) \\ $$$$\Rightarrow\:\:\mathrm{BF}^{\mathrm{2}} =\mathrm{2}\boldsymbol{\mathrm{R}}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}\boldsymbol{\alpha}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\bullet\mathrm{triangle}\:\left(\mathrm{ABE}\right)\:\:\mathrm{AE}\bot\mathrm{BE}\:\:\:\:\left(\mathrm{BE}=\boldsymbol{\mathrm{r}}\right) \\ $$$$\mathrm{AB}^{\mathrm{2}} =\mathrm{AE}^{\mathrm{2}} +\boldsymbol{\mathrm{r}}^{\mathrm{2}} \Rightarrow\:\:\:\:\:\mathrm{2}\boldsymbol{\mathrm{R}}^{\mathrm{2}} =\mathrm{16}^{\mathrm{2}} +\boldsymbol{\mathrm{r}}^{\mathrm{2}} \\ $$$$\bullet\mathrm{BEF}\:\:\:\mathrm{BF}^{\mathrm{2}} =\mathrm{BE}^{\mathrm{2}} +\mathrm{EF}^{\mathrm{2}} =\boldsymbol{\mathrm{r}}^{\mathrm{2}} +\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\boldsymbol{\mathrm{R}}^{\mathrm{2}} −\mathrm{231}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\:\:\:\mathrm{2}\boldsymbol{\mathrm{R}}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{sin}\:\mathrm{2}\boldsymbol{\alpha}\right)=\mathrm{2}\boldsymbol{\mathrm{R}}^{\mathrm{2}} −\mathrm{231} \\ $$$$\:\:\:\mathrm{2}\boldsymbol{\mathrm{R}}^{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\boldsymbol{\alpha}−\mathrm{231}=\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\mathrm{2}\left(\frac{\mathrm{11}^{\mathrm{2}} }{\mathrm{4cos}\:^{\mathrm{2}} \boldsymbol{\alpha}}\right)\mathrm{sin}\:\mathrm{2}\boldsymbol{\alpha}=\mathrm{231} \\ $$$$\left(\mathrm{sin}\:\mathrm{2}\alpha=\frac{\mathrm{2}\boldsymbol{\mathrm{t}}}{\mathrm{1}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} },\:\:\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\alpha}}=\mathrm{1}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} ;\:\:\boldsymbol{\mathrm{t}}=\mathrm{tan}\:\alpha\right) \\ $$$$\:\:\:\:\:\:\mathrm{121}\boldsymbol{\mathrm{t}}=\mathrm{231}\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{t}}=\frac{\mathrm{231}}{\mathrm{121}}\:\:\:\:\:\:\:\left(\mathrm{5}\right) \\ $$$$\boldsymbol{\mathrm{dans}}\:\boldsymbol{\mathrm{l}}\:\boldsymbol{\mathrm{expression}}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{S}}=\mathrm{44tan}\:\boldsymbol{\alpha}=\frac{\mathrm{924}}{\mathrm{11}} \\ $$$$\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{Surface}}\:\boldsymbol{\mathrm{Triangle}}\:\left(\boldsymbol{\mathrm{AOE}}\right)\:\:=\mathrm{84} \\ $$$$\: \\ $$
Commented by a.lgnaoui last updated on 17/Jun/23
Commented by Lekhraj last updated on 26/Jun/23
Thank a lot for this wonderfull solution.
$${Thank}\:{a}\:{lot}\:{for}\:{this}\:{wonderfull}\:{solution}.\: \\ $$

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