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Question Number 191455 by Mingma last updated on 24/Apr/23

Answered by a.lgnaoui last updated on 25/Apr/23

$$△\mathrm{ANB}\mathrm{BM}\mathrm{\perp }\mathrm{AN}\mathrm{et}\measuredangle \mathrm{MBA}=\measuredangle \mathrm{MBN}$$$$\Rightarrow \mathrm{AB}=\mathrm{BN}=\frac{\mathbf{x}}{\cos \mathit{\theta }}\left(1\right)$$$$△\mathrm{PNC}\mathrm{PC}=3\mathrm{cm}$$$$\mathrm{BM}\mid \mid \mathrm{NP}\Rightarrow \measuredangle \mathrm{PNC}=\measuredangle \mathrm{MBN}=\theta =\measuredangle \mathrm{PCN}$$$$\Rightarrow \mathrm{NP}=\mathrm{PC}=3$$$$\mathrm{NC}=2\mathrm{PCcos}\theta =6\cos \theta$$$$△\mathrm{ABC}\measuredangle \mathrm{ABN}=2\theta$$$${\mathrm{BC}}^2={\mathrm{AC}}^2+{\mathrm{AB}}^2+2\mathrm{AC}\times \mathrm{ABcos}3\theta$$$$$$$${\mathbf{BC}}^2=100+\frac{{\mathbf{x}}^2}{\cos {}^2\mathit{\theta }}+\frac{20\mathbf{x}}{\cos \mathit{\theta }}\cos 3\theta \left(2\right)$$$$\mathrm{Calcul}\mathrm{de}\mathrm{BC}:$$$$\mathrm{BC}=\mathrm{BN}+\mathrm{NC}=\frac{\mathrm{x}}{\cos \theta }+6\cos \theta$$$$\Rightarrow {\mathrm{BC}}^2=\frac{{\mathbf{x}}^2}{\cos {}^2\mathit{\theta }}+36\cos {}^2\mathit{\theta }+12\mathbf{x}\left(3\right)$$$$\left(2\right)=\left(3\right)\Rightarrow \frac{20\mathrm{xcos}3\theta }{\cos \theta }+100=36\cos {}^2\theta +12\mathrm{x}$$$$\frac{20\mathrm{x}(4\cos {}^2\theta -3)\cos \theta }{\cos \theta }+100-12\mathrm{x}-36\cos {}^2\theta =0$$$$[20(4\cos {}^2\theta -3)-12]\mathrm{x}={36\cos }^2\theta -100$$$$[5(4\cos {}^2\theta -3)-3]\mathrm{x}=9\cos {}^2\theta -25$$$$\mathbf{x}=\frac{9{\mathbf{cos}}^2\mathit{\theta }-25}{20\mathbf{cos}{}^2\mathit{\theta }-15}\left(4\right)$$$$△\mathrm{ANC}\mathrm{AN}=2\mathrm{MN}=2\mathrm{xtan}\theta$$$${\mathrm{AC}}^2={\mathrm{AN}}^2+{\mathrm{NC}}^2-2\mathrm{AN}\times \mathrm{NCcos}(\frac{\pi }{2}+\theta )$$$$\Rightarrow 100={4\mathrm{x}}^2\tan {}^2\theta +36\cos {}^2\theta +$$$$4\mathrm{xtan}\theta \times 6\cos \theta \sin \theta$$$$={4\mathrm{x}}^2\tan {}^2\theta +36\cos {}^2\theta +24\mathrm{xsin}{}^2\theta$$$$25={\mathrm{x}}^2\tan {}^2\theta +{6\mathrm{xsin}}^2\theta +9\cos {}^2\theta$$$$25\cos {}^2\theta ={\mathrm{x}}^2\sin {}^2\theta +\mathrm{x}6\sin {}^2\theta \cos {}^2\theta +9\cos {}^4\theta$$$$$$$${\mathrm{x}}^2+{6\mathrm{xcos}}^2\theta +\frac{9}{(1-\cos {}^2\theta )}\cos {}^4\theta -25\left(\frac{\cos {}^2\theta }{1-\cos {}^2\theta }\right)=0\left(5\right)$$$$$$$$\left(4\right)\iff (20\cos {}^2\theta -15)\mathrm{x}=9\cos {}^2\theta -25$$$$$$$$\Rightarrow (20\mathbf{x}-9){\cos }^2\mathit{\theta }=15\mathrm{x}-25$$$$\cos {}^2\theta =\frac{5(3\mathrm{x}-5)}{20\mathrm{x}-9}$$$$$$$${\mathrm{x}}^2+6\times \frac{5(3\mathrm{x}-5)}{20\mathrm{x}-9}+\frac{9{\left(\frac{5(3\mathrm{x}-5)}{20\mathrm{x}-9}\right)}^2}{1-\frac{5(3\mathrm{x}-5)}{(20\mathrm{x}-9)}}$$$$-25\times \frac{\frac{5(3\mathrm{x}-5)}{20\mathrm{x}-9}}{\frac{(20\mathrm{x}-9)-5(3\mathrm{x}-5)}{20\mathrm{x}-9}}$$$$\iff {\mathrm{x}}^2+\frac{30(3\mathrm{x}-5)}{20\mathrm{x}-9}+\frac{25\times 9{(3\mathrm{x}-5)}^2}{(20\mathrm{x}-9)(5\mathrm{x}+16)}$$$$-\frac{25\times 5(3\mathrm{x}-5)}{5\mathrm{x}+16}$$$${\mathbf{x}}^2(20\mathbf{x}-9)(5\mathbf{x}+16)+30(3\mathbf{x}-5)(5\mathbf{x}+16)$$$$+225{(3\mathbf{x}-5)}^2-125(3\mathbf{x}-5)(20\mathbf{x}-9)=0$$$$$$$${\mathbf{x}}^2(100{\mathbf{x}}^2+275\mathbf{x}-144)+30(15{\mathbf{x}}^2+13\mathbf{x}-80)$$$$+225(9{\mathbf{x}}^2-30\mathbf{x}+25)-125(60{\mathbf{x}}^2-127\mathbf{x}+45)$$$$$$$$100{\mathbf{x}}^4+275{\mathbf{x}}^3-5269{\mathbf{x}}^2+9515\mathbf{x}-2400=0$$$$$$$$\mathrm{Resolution}\mathrm{d}\mathrm{equation}\mathrm{donne}$$$$$$$$\mathrm{valeur}\mathrm{accepte}\mathbf{x}=4,513\mathbf{cm}$$$$$$$$$$

Commented by mr W last updated on 25/Apr/23

$$wrong!$$

Answered by mr W last updated on 25/Apr/23

Commented by Mingma last updated on 26/Apr/23

Excellent, sir!

Commented by mr W last updated on 25/Apr/23

$$NP=PC=3$$$$MQ=\frac{NP}{2}=1.5$$$$QP=\frac{AP}{2}=3.5$$$$BQ=QC$$$$x+1.5=3.5+3$$$$\Rightarrow x=5✓$$

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