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Question Number 192819 by cherokeesay last updated on 28/May/23

Answered by a.lgnaoui last updated on 28/May/23

$$△\mathrm{PBM}\mathrm{recrangle}\mathrm{en}\mathrm{B}$$$$\measuredangle \mathrm{PMB}=\lambda$$$$△\mathrm{MAD}\mathrm{MA}=\frac{1}{2}\mathrm{AD}\Rightarrow \measuredangle \mathrm{AMD}=\measuredangle \mathrm{MDC}=\frac{\pi }{3}$$$$\mathrm{MD}=\sqrt{{\mathrm{AD}}^2+{\left(\frac{\mathrm{AD}}{2}\right)}^2}=\frac{\mathrm{AD}\sqrt{5}}{2}(\mathrm{AD}=\mathrm{AB})$$$$△\mathrm{PMB}\mathrm{et}△\mathrm{PNC}\mathrm{semblables}$$$$\Rightarrow \measuredangle \mathrm{PNC}=\measuredangle \mathrm{PMB}=\lambda$$$$\mathrm{de}\mathrm{plus}\lambda =\theta +\mathrm{x}(\theta =\measuredangle \mathrm{DPM})$$$$\tan \lambda =\frac{\mathrm{PB}}{\mathrm{MB}}=\frac{\mathrm{BC}+\mathrm{PC}}{\mathrm{BC}/2}$$$$△\mathrm{DPC}\frac{\sin \mathrm{x}}{\mathrm{PC}}=\frac{\cos \mathrm{x}}{\mathrm{CD}}\Rightarrow \mathrm{PC}=\mathrm{CDtan}\mathrm{x}=\mathrm{BCtan}\mathrm{x}$$$$\Rightarrow \tan \lambda =\frac{2\mathrm{BC}(1+\tan \mathrm{x})}{\mathrm{BC}}=2(1+\tan \mathrm{x})\left(1\right)$$$$$$$$△\mathrm{DNP}\frac{\sin (\mathrm{x}+\frac{\pi }{3})}{\mathrm{MP}}=\frac{\sin \theta }{\mathrm{MD}}=\frac{2\sin (\lambda -\mathrm{x})}{\mathrm{AD}\sqrt{5}}\left(2\right)$$$$$$$$\cos \lambda =\frac{\mathrm{BC}}{2\mathrm{MP}}\mathrm{MP}=\frac{\mathrm{BC}}{2\cos \lambda }$$$$\Rightarrow$$$$\frac{2\sin (\mathrm{x}+\frac{\pi }{3})\cos \lambda }{\mathrm{BC}}=\frac{2\sin (\lambda -\mathrm{x})}{\mathrm{BC}\sqrt{5}}$$$$$$$$\iff \sin (\mathrm{x}+\frac{\pi }{3})\cos \lambda =\frac{\sqrt{5}\sin (\lambda -\mathrm{x})}{5}$$$$\frac{1}{2}(\sin \mathrm{x}+\sqrt{3}\cos \mathrm{x})\cos \lambda =\frac{\sqrt{5}}{5}(\sin \lambda \cos \mathrm{x}-\cos \lambda \sin \mathrm{x})$$$$\frac{1}{2}(\tan \mathrm{x}+\sqrt{3})\cos \lambda =\frac{\sqrt{5}}{5}(\sin \lambda -\cos \lambda \tan \mathrm{x})$$$$\iff \frac{1}{2}(\tan \mathrm{x}+\sqrt{3})=\frac{\sqrt{5}}{5}(\tan \lambda -\tan \mathrm{x})$$$$$$$$\mathrm{d}\mathrm{apres}\left(2\right)\tan \lambda =2(1+\tan \mathrm{x})$$$$$$$$\frac{1}{2}(\mathbf{tan}\mathbf{x}+\sqrt{3})=\frac{\sqrt{5}}{5}(2+\mathbf{tan}\mathbf{x})$$$$$$$$\mathbf{tan}\mathbf{x}=\frac{\frac{4\sqrt{5}}{5}-\sqrt{3}}{1-\frac{2\sqrt{5}}{5}}=\frac{4\sqrt{5}-5\sqrt{3}}{5-2\sqrt{5}}$$$$\mathbf{x}={\tan }^{-1}\left(\frac{4\sqrt{5}-5\sqrt{3}}{5-2\sqrt{5}}\right)$$$$$$$$\mathbf{soit}:\mathbf{x}=28,28°$$$$$$

Commented by a.lgnaoui last updated on 28/May/23

Commented by cherokeesay last updated on 28/May/23

$$30°toutjuste!$$

Answered by cherokeesay last updated on 28/May/23

Commented by MM42 last updated on 29/May/23

$$verynice$$

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