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Question Number 197914 by a.lgnaoui last updated on 04/Oct/23

$$\mathbf{Determiner}\mathbf{la}\mathbf{surface}\mathbf{hachuree}$$$$\left(\mathbf{voir}\mathbf{figure}\right)$$$$\mathbf{BC}=10\mathbf{cm}\measuredangle \mathbf{B}=45°\measuredangle \mathbf{C}=30°$$

Commented by a.lgnaoui last updated on 04/Oct/23

Commented by a.lgnaoui last updated on 05/Oct/23

$$Reponse:$$$$\mathbf{AB}\sin 45°=\mathbf{AC}\sin 30°$$$$\frac{\sqrt{2}}{2}\mathbf{AB}=\frac{1}{2}\mathbf{AC}\Rightarrow \mathbf{AC}=\mathbf{AB}\sqrt{2}$$$${\mathbf{BC}}^2={\mathbf{AB}}^2+{\mathbf{AC}}^2-2\mathbf{AB}\times \mathbf{AC}\cos 105°$$$$100={3\mathrm{AB}}^2-2\sqrt{2}{\mathrm{AB}}^2\cos 105$$$$=(3-2\sqrt{2}\cos 105){\mathrm{AB}}^2$$$$\mathbf{AB}=\frac{10}{\sqrt{3-2\sqrt{2}\cos 105}}$$$$\mathbf{AB}=5,176$$$$\bullet △\mathrm{OBE}\sin 45°=\frac{\mathbf{OE}}{\mathbf{OB}}=\frac{\mathbf{R}1}{\mathbf{AB}-\mathbf{R}1}$$$$\Rightarrow (\mathbf{AB}-\mathbf{R}1)\frac{\sqrt{2}}{2}=\mathbf{R}1$$$$\mathbf{AB}-\mathbf{R}1=\mathbf{R}1\sqrt{2}\mathbf{R}1=\frac{\mathbf{AB}}{1+\sqrt{2}}$$$$\mathbf{R}1=\frac{10}{(1+\sqrt{2})\sqrt{(3-2\sqrt{2}\cos 105)}}$$$$$$$$\mathbf{AN}\mathbf{R}1=2,14$$$$\bullet \measuredangle \mathbf{OFC}\sin 30=\frac{\mathbf{R}2}{\mathbf{OC}}=\frac{\mathbf{R}2}{\mathbf{AC}-\mathbf{R}2}$$$$\mathbf{AC}=\mathbf{AB}\sqrt{2}\Rightarrow \mathbf{AC}=5,176\times \sqrt{2}$$$$\mathbf{AC}=7,32$$$$\frac{\mathbf{AC}}{2}-\frac{\mathbf{R}2}{2}=\mathbf{R}2\mathbf{R}2=\frac{\mathbf{AC}}{3}=2,44$$$$\bullet \mathbf{Calcul}\mathbf{de}\mathbf{la}\mathbf{surface}$$$$\mathbf{S}=\mathbf{S}1+\mathbf{S}2$$$$\mathbf{S}1=\frac{\mathit{\pi }\mathbf{R}{1}^2\times \mathit{\alpha }}{360}-\mathbf{S}(△\mathbf{OAM})$$$$\mathbf{S}2=\frac{\mathit{\pi }\mathbf{R}{2}^2\times \mathit{\beta }}{360}-\mathbf{S}\left({\mathbf{AO}}^{\prime }\mathbf{M}\right)$$$$△\mathbf{AOM}=\frac{\sin \mathbf{A}1}{\mathbf{R}1}=\frac{\sin 2\mathbf{A}1}{\mathbf{AM}}$$$$\mathrm{A}1=\alpha \Rightarrow \frac{\sin \alpha }{\mathbf{R}1}=\frac{\sin 2\mathit{\alpha }}{\mathbf{AM}}$$$$\mathbf{AM}=2\mathbf{R}1\cos \mathit{\alpha }$$$$\mathbf{S}\left(\mathbf{OSM}\right)=\frac{\mathit{\pi }\mathbf{R}{1}^2}{360}\mathit{\alpha }-\frac{\mathrm{AM}}{2}\mathrm{R}1\cos \mathit{\alpha }$$$$=\frac{\pi {\mathrm{R}1}^2\alpha }{360}-\mathbf{R}{1}^2\cos {}^2\mathit{\alpha }$$$$\mathbf{S}\left({\mathbf{AO}}^{\prime }\mathbf{M}\right)=\frac{\mathit{\pi }\mathbf{R}2\times (105-\alpha )}{360}-\mathbf{R}{2}^2{\cos }^2(105-\alpha )$$$$(\beta =105-\alpha )$$$$\left(\mathit{A}\mathit{S}\mathit{u}\mathit{i}\mathit{v}\mathit{r}\mathit{e}\right)...........$$$$$$$$△\mathbf{AOM}/{\mathbf{AO}}^{\prime }\mathbf{M}$$$${\mathbf{OM}}^2={\mathbf{AO}}^2+{\mathbf{AM}}^2-2\mathbf{AO}\times \mathbf{AM}\cos \mathit{\alpha }$$$$\mathbf{R}{1}^2=\mathbf{R}{1}^2+{\mathbf{AM}}^2-2\mathbf{AM}\times \mathbf{R}1\cos \mathit{\alpha }$$$$0={\mathbf{AM}}^2-2\mathbf{AM}\times \mathbf{R}1\cos \mathit{\alpha }$$$$$$$$\mathbf{AM}=2\mathbf{R}1\cos \mathit{\alpha }\left(\mathbf{i}\right)$$$$$$$$\mathrm{de}\mathrm{meme}$$$$\mathbf{AM}=2\mathbf{R}2\cos (105-\mathit{\alpha })$$$$\Rightarrow \mathbf{R}1\cos \mathit{\alpha }=\mathbf{R}2\cos (105-\mathit{\alpha })$$$$\frac{\cos \mathit{\alpha }}{\cos (105-\mathit{\alpha })}=\frac{\mathbf{R}2}{\mathbf{R}1}=\frac{122}{107}$$$$122(\cos 105\cos \alpha +\sin 105\sin \alpha )=107\cos \alpha$$$$(122\cos 105-107)\cos \mathit{\alpha }+122\times \sin 105\sin \mathit{\alpha }=0$$$$(107-122\cos 105)\cos \mathit{\alpha }=122\sin \mathit{\alpha }$$$$\Rightarrow \tan \mathit{\alpha }=\frac{107-122\cos 105}{122}=1,135$$$$\measuredangle \mathbf{OAM}=48,6°$$$$\mathbf{AM}=4,28\cos \mathit{\alpha }=2,83$$$$\Rightarrow \measuredangle \mathbf{AOH}\approx 90-48=42°$$$$\mathbf{S}1=\frac{\mathit{\pi }{\mathbf{R}}^{2}1\times 84}{360}-\left(\frac{2,83}{2}\right)\times \mathbf{R}1\cos 42$$$$\mathbf{S}1=3,357-2,25=1,106$$$$$$$$\mathbf{S}2=\frac{\mathit{\pi }\mathbf{R}{2}^2\times 57}{360}-\left(\frac{2,83}{2}\right)\times \mathbf{R}2\cos 33$$$$=2,961-2,895=0,066$$$$$$$$\mathbf{Surface}\mathbf{est}1,106+0,066$$$$\mathbf{S}=1,172$$$$$$

Commented by a.lgnaoui last updated on 05/Oct/23

Answered by a.lgnaoui last updated on 06/Oct/23

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