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Question Number 202093 by sonukgindia last updated on 20/Dec/23

Commented by a.lgnaoui last updated on 22/Dec/23

Commented by a.lgnaoui last updated on 22/Dec/23

$$\mathrm{\Delta }\mathrm{AMN}\measuredangle \mathrm{MNA}=\frac{3\pi }{4}-\alpha$$$$\measuredangle \mathrm{ANM}=\frac{\pi }{4}+\alpha \Rightarrow \measuredangle \mathrm{ACE}=\alpha$$$$\mathbf{Calcul}\mathbf{de}\mathbf{AB}\mathbf{AC}\mathbf{en}\mathbf{finction}\mathbf{de}\mathbf{x};\mathit{\alpha }:$$$$\measuredangle \mathrm{ABC}\frac{\sin (45+\alpha )}{\mathrm{AC}}=\frac{\cos \alpha }{\mathrm{AB}}=\frac{\sin 45}{\sqrt{2}}$$$$\mathrm{AC}=\frac{\sqrt{2}\sin (45+\alpha )}{\sin 45}=\sqrt{2}(\sin \alpha +\cos \alpha )$$$$\mathbf{AC}=\sqrt{2}(\sin \mathit{\alpha }+\cos \mathit{\alpha })\left(1\right)$$$$\mathrm{AB}=\frac{\sqrt{2}\cos \alpha }{\sin 45}=2\cos \alpha$$$$\mathbf{AB}=2\cos \mathit{\alpha }\left(2\right)$$$$$$$$\mathrm{Camcul}\mathrm{de}\mathrm{AB}$$$$\mathrm{AB}=\mathrm{AM}+\mathrm{MB}$$$$\frac{\sin 45}{\mathrm{x}}=\frac{\sin (\frac{3\pi }{4}-\alpha )}{\mathrm{AM}};\mathrm{AM}=\frac{\mathrm{xsin}(\frac{3\pi }{4}-\alpha )}{\sin \frac{\pi }{4}}$$$$\Rightarrow \mathbf{AM}=\mathbf{x}(\cos \mathit{\alpha }+\sin \mathit{\alpha })\left(3\right)$$$$\mathrm{\Delta }\mathrm{MBD}\frac{\sin \frac{\pi }{4}}{\mathrm{MB}}=\frac{\sin \alpha }{\mathrm{a}}$$$$\Rightarrow \mathbf{MB}=\frac{\mathbf{a}\sin 45}{\sin \mathit{\alpha }}=\frac{\mathbf{a}\sqrt{2}}{2\sin \mathit{\alpha }}\left(4\right)$$$$\mathrm{akora}\mathbf{AB}=\mathbf{AM}+\mathbf{MB}$$$$\mathbf{AB}=\mathbf{x}(\sin \mathit{\alpha }+\cos \mathit{\alpha })+\frac{\mathrm{a}\sqrt{2}}{2\sin \mathit{\alpha }}\left(5\right)$$$$\bullet \mathrm{Calcul}\mathrm{de}\mathbf{AC}$$$$\mathrm{AC}=\mathrm{AN}+\mathrm{NC}$$$$\frac{\sin \alpha }{\mathrm{AN}}=\frac{\sin (\frac{3\pi }{4}-\alpha )}{\mathrm{AM}}=\frac{\sqrt{2}}{2\mathrm{x}}$$$$\Rightarrow \mathbf{AN}=\sqrt{2}\mathbf{x}\sin \mathit{\alpha }\left(6\right)$$$$\bullet \mathrm{Calcul}\mathrm{de}\mathrm{ND}$$$$\measuredangle \mathrm{ANE}\frac{\cos \alpha }{\mathrm{NE}}=\frac{\sin 45}{\mathrm{AN}}$$$$\Rightarrow \mathrm{NE}=\frac{\mathrm{ANcos}\alpha }{\sin 45}=\frac{2\mathrm{xsin}\alpha \cos \alpha }{}$$$$\mathrm{NE}=\mathrm{xsin}2\alpha$$$$\mathrm{\Delta }\mathrm{NEC}\frac{\sin \alpha }{\mathrm{NE}}=\frac{\sin 45}{\mathrm{ND}}$$$$\Rightarrow \mathrm{ND}=\frac{\mathbf{x}\sin 2\mathit{\alpha }\sin 45}{\sin \alpha }=\frac{2\sqrt{2}\mathrm{xcos}\alpha }{2}$$$$\mathbf{ND}=\sqrt{2}\mathbf{x}\cos \mathit{\alpha }$$$$\mathrm{alors}\mathrm{AC}=\sqrt{2}\mathbf{x}\sin \mathit{\alpha }+\sqrt{2}\mathbf{x}\cos \alpha$$$$$$$$\mathbf{AC}=\mathbf{x}\sqrt{2}(\sin \mathit{\alpha }+\cos \mathit{\alpha })\left(7\right)$$$$\left(1\right)\left(2\right)\Rightarrow$$$$\{\begin{array}{l}\frac{\mathbf{a}\sqrt{2}}{2\sin \mathit{\alpha }}+\mathbf{x}(\sin \mathit{\alpha }+\cos \mathit{\alpha })=2\cos \mathit{\alpha }\left(8\right)\\ \mathbf{x}\sqrt{2}(\sin \mathit{\alpha }+\cos \mathit{\alpha })=\sqrt{2}(\cos \mathit{\alpha }+\sin \mathit{\alpha })\left(9\right)\end{array}$$$$\left(9\right)\Rightarrow \mathbf{x}\sqrt{2}=\sqrt{2}$$$$$$$$\mathbf{donc}\mathbf{x}=1$$$$$$$$$$

Answered by MathematicalUser2357 last updated on 22/Dec/23

$$\mathrm{Is}\mathrm{the}\mathrm{size}\mathrm{of}\mathrm{the}\mathrm{yellow}\mathrm{angle}45°?$$

Commented by a.lgnaoui last updated on 22/Dec/23

$$\mathrm{from}\mathrm{picture}\mathrm{yellow}\mathrm{angle}=45°$$

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