Question Number 191455 by Mingma last updated on 24/Apr/23
Answered by a.lgnaoui last updated on 25/Apr/23
![△ANB BM⊥AN et ∡MBA=∡MBN ⇒AB=BN=(x/(cos 𝛉)) (1) △PNC PC=3cm BM∣∣ NP ⇒ ∡ PNC=∡MBN=θ=∡PCN ⇒NP=PC=3 NC=2PCcos θ=6cos θ △ABC ∡ABN=2θ BC^2 =AC^2 +AB^2 +2AC×ABcos 3θ BC^2 =100+(x^2 /(cos^2 𝛉))+((20x)/(cos 𝛉))cos 3θ (2) Calcul de BC: BC=BN+NC=(x/(cos θ))+6cos θ ⇒BC^2 =(x^2 /(cos^2 𝛉))+36cos^2 𝛉+12x (3) (2)=(3)⇒ ((20xcos 3θ)/(cos θ))+100=36cos^2 θ+12x ((20x(4cos^2 θ−3)cos θ)/(cos θ))+100−12x−36cos^2 θ=0 [20(4cos^2 θ−3)−12]x=36cos^2 θ −100 [5(4cos^2 θ−3)−3]x=9cos^2 θ−25 x=((9cos^2 𝛉−25)/(20cos^2 𝛉−15)) (4) △ANC AN=2MN=2xtan θ AC^2 =AN^2 +NC^2 −2AN×NCcos ((π/2)+θ) ⇒100=4x^2 tan^2 θ+36cos^2 θ+ 4xtan θ×6cos θsin θ =4x^2 tan^2 θ+36cos^2 θ+24xsin^2 θ 25=x^2 tan^2 θ+6xsin^2 θ+9cos^2 θ 25cos^2 θ=x^2 sin^2 θ+x6sin^2 θcos^2 θ+9cos^4 θ x^2 +6xcos^2 θ+ (9/((1−cos ^2 θ)))cos^4 θ−25(((cos^2 θ)/(1−cos^2 θ)))=0 (5) (4)⇔(20cos^2 θ−15)x=9cos^2 θ−25 ⇒ (20x−9)cos^2 𝛉 =15x−25 cos^2 θ=((5(3x−5))/(20x−9)) x^2 +6×((5(3x−5))/(20x−9))+((9(((5(3x−5))/(20x−9)))^2 )/(1−((5(3x−5))/((20x−9))))) −25×(((5(3x−5))/(20x−9))/(((20x−9)−5(3x−5))/(20x−9))) ⇔ x^2 +((30(3x−5))/(20x−9))+((25×9(3x−5)^2 )/((20x−9)(5x+16))) −((25×5(3x−5))/(5x+16)) x^2 (20x−9)(5x+16)+30(3x−5)(5x+16) +225(3x−5)^2 −125(3x−5)(20x−9)=0 x^2 (100x^2 +275x−144)+30(15x^2 +13x−80) +225(9x^2 −30x+25)−125(60x^2 −127x+45) 100x^4 +275x^3 −5269x^2 +9515x−2400=0 Resolution d equation donne valeur accepte x=4,513 cm](https://www.tinkutara.com/question/Q191551.png)
$$\:\bigtriangleup\mathrm{ANB}\:\:\:\mathrm{BM}\bot\mathrm{AN}\:\:\mathrm{et}\:\measuredangle\mathrm{MBA}=\measuredangle\mathrm{MBN} \\ $$$$\:\Rightarrow\mathrm{AB}=\mathrm{BN}=\frac{\boldsymbol{\mathrm{x}}}{\mathrm{cos}\:\boldsymbol{\theta}}\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\:\bigtriangleup\mathrm{PNC}\:\:\:\:\:\:\mathrm{PC}=\mathrm{3cm} \\ $$$$\:\mathrm{BM}\mid\mid\:\mathrm{NP}\:\:\Rightarrow\:\measuredangle\:\mathrm{PNC}=\measuredangle\mathrm{MBN}=\theta=\measuredangle\mathrm{PCN} \\ $$$$\:\Rightarrow\mathrm{NP}=\mathrm{PC}=\mathrm{3} \\ $$$$\:\mathrm{NC}=\mathrm{2PCcos}\:\theta=\mathrm{6cos}\:\theta \\ $$$$\:\bigtriangleup\mathrm{ABC}\:\:\:\measuredangle\mathrm{ABN}=\mathrm{2}\theta \\ $$$$\:\mathrm{BC}^{\mathrm{2}} =\mathrm{AC}^{\mathrm{2}} +\mathrm{AB}^{\mathrm{2}} +\mathrm{2AC}×\mathrm{ABcos}\:\mathrm{3}\theta \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{BC}}^{\mathrm{2}} \:\:\:\:\:\:\:\:=\mathrm{100}+\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\theta}}+\frac{\mathrm{20}\boldsymbol{\mathrm{x}}}{\mathrm{cos}\:\boldsymbol{\theta}}\mathrm{cos}\:\mathrm{3}\theta\:\left(\mathrm{2}\right) \\ $$$$\:\:\mathrm{Calcul}\:\:\mathrm{de}\:\mathrm{BC}: \\ $$$$\:\:\mathrm{BC}=\mathrm{BN}+\mathrm{NC}=\frac{\mathrm{x}}{\mathrm{cos}\:\theta}+\mathrm{6cos}\:\theta \\ $$$$\Rightarrow\mathrm{BC}^{\mathrm{2}} =\frac{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{cos}\:^{\mathrm{2}} \boldsymbol{\theta}}+\mathrm{36cos}\:^{\mathrm{2}} \boldsymbol{\theta}+\mathrm{12}\boldsymbol{\mathrm{x}}\:\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)=\left(\mathrm{3}\right)\Rightarrow\:\frac{\mathrm{20xcos}\:\mathrm{3}\theta}{\mathrm{cos}\:\theta}+\mathrm{100}=\mathrm{36cos}\:^{\mathrm{2}} \theta+\mathrm{12x} \\ $$$$\frac{\mathrm{20x}\left(\mathrm{4cos}\:^{\mathrm{2}} \theta−\mathrm{3}\right)\mathrm{cos}\:\theta}{\mathrm{cos}\:\theta}+\mathrm{100}−\mathrm{12x}−\mathrm{36cos}\:^{\mathrm{2}} \theta=\mathrm{0} \\ $$$$\left[\mathrm{20}\left(\mathrm{4cos}\:^{\mathrm{2}} \theta−\mathrm{3}\right)−\mathrm{12}\right]\mathrm{x}=\mathrm{36cos}^{\mathrm{2}} \theta\:−\mathrm{100} \\ $$$$\left[\mathrm{5}\left(\mathrm{4cos}\:^{\mathrm{2}} \theta−\mathrm{3}\right)−\mathrm{3}\right]\mathrm{x}=\mathrm{9cos}\:^{\mathrm{2}} \theta−\mathrm{25} \\ $$$$\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\frac{\mathrm{9}\boldsymbol{\mathrm{cos}}^{\mathrm{2}} \boldsymbol{\theta}−\mathrm{25}}{\mathrm{20}\boldsymbol{\mathrm{cos}}\:^{\mathrm{2}} \boldsymbol{\theta}−\mathrm{15}}\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$$\bigtriangleup\mathrm{ANC}\:\:\:\:\mathrm{AN}=\mathrm{2MN}=\mathrm{2xtan}\:\theta \\ $$$$\:\:\mathrm{AC}^{\mathrm{2}} =\mathrm{AN}^{\mathrm{2}} +\mathrm{NC}^{\mathrm{2}} −\mathrm{2AN}×\mathrm{NCcos}\:\left(\frac{\pi}{\mathrm{2}}+\theta\right) \\ $$$$\Rightarrow\mathrm{100}=\mathrm{4x}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{36cos}\:^{\mathrm{2}} \theta+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4xtan}\:\theta×\mathrm{6cos}\:\theta\mathrm{sin}\:\theta \\ $$$$\:\:\:\:\:=\mathrm{4x}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{36cos}\:^{\mathrm{2}} \theta+\mathrm{24xsin}\:^{\mathrm{2}} \theta \\ $$$$\mathrm{25}=\mathrm{x}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta+\mathrm{6xsin}^{\mathrm{2}} \theta+\mathrm{9cos}\:^{\mathrm{2}} \theta \\ $$$$\mathrm{25cos}\:^{\mathrm{2}} \theta=\mathrm{x}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta+\mathrm{x6sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{9cos}\:^{\mathrm{4}} \theta \\ $$$$ \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{6xcos}^{\mathrm{2}} \theta+\:\frac{\mathrm{9}}{\left(\mathrm{1}−\mathrm{cos}\:\:^{\mathrm{2}} \theta\right)}\mathrm{cos}\:^{\mathrm{4}} \theta−\mathrm{25}\left(\frac{\mathrm{cos}\:^{\mathrm{2}} \theta}{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \theta}\right)=\mathrm{0}\:\:\left(\mathrm{5}\right) \\ $$$$ \\ $$$$\left(\mathrm{4}\right)\Leftrightarrow\left(\mathrm{20cos}\:^{\mathrm{2}} \theta−\mathrm{15}\right)\mathrm{x}=\mathrm{9cos}\:^{\mathrm{2}} \theta−\mathrm{25} \\ $$$$ \\ $$$$\:\:\:\:\:\Rightarrow\:\:\left(\mathrm{20}\boldsymbol{\mathrm{x}}−\mathrm{9}\right)\mathrm{cos}^{\mathrm{2}} \boldsymbol{\theta}\:=\mathrm{15x}−\mathrm{25} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{cos}\:^{\mathrm{2}} \theta=\frac{\mathrm{5}\left(\mathrm{3x}−\mathrm{5}\right)}{\mathrm{20x}−\mathrm{9}} \\ $$$$ \\ $$$$\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{6}×\frac{\mathrm{5}\left(\mathrm{3x}−\mathrm{5}\right)}{\mathrm{20x}−\mathrm{9}}+\frac{\mathrm{9}\left(\frac{\mathrm{5}\left(\mathrm{3x}−\mathrm{5}\right)}{\mathrm{20x}−\mathrm{9}}\right)^{\mathrm{2}} }{\mathrm{1}−\frac{\mathrm{5}\left(\mathrm{3x}−\mathrm{5}\right)}{\left(\mathrm{20x}−\mathrm{9}\right)}} \\ $$$$−\mathrm{25}×\frac{\frac{\mathrm{5}\left(\mathrm{3x}−\mathrm{5}\right)}{\mathrm{20x}−\mathrm{9}}}{\frac{\left(\mathrm{20x}−\mathrm{9}\right)−\mathrm{5}\left(\mathrm{3x}−\mathrm{5}\right)}{\mathrm{20x}−\mathrm{9}}} \\ $$$$\Leftrightarrow\:\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{30}\left(\mathrm{3x}−\mathrm{5}\right)}{\mathrm{20x}−\mathrm{9}}+\frac{\mathrm{25}×\mathrm{9}\left(\mathrm{3x}−\mathrm{5}\right)^{\mathrm{2}} }{\left(\mathrm{20x}−\mathrm{9}\right)\left(\mathrm{5x}+\mathrm{16}\right)} \\ $$$$\:\:\:\:\:−\frac{\mathrm{25}×\mathrm{5}\left(\mathrm{3x}−\mathrm{5}\right)}{\mathrm{5x}+\mathrm{16}} \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} \left(\mathrm{20}\boldsymbol{\mathrm{x}}−\mathrm{9}\right)\left(\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{16}\right)+\mathrm{30}\left(\mathrm{3}\boldsymbol{\mathrm{x}}−\mathrm{5}\right)\left(\mathrm{5}\boldsymbol{\mathrm{x}}+\mathrm{16}\right) \\ $$$$+\mathrm{225}\left(\mathrm{3}\boldsymbol{\mathrm{x}}−\mathrm{5}\right)^{\mathrm{2}} −\mathrm{125}\left(\mathrm{3}\boldsymbol{\mathrm{x}}−\mathrm{5}\right)\left(\mathrm{20}\boldsymbol{\mathrm{x}}−\mathrm{9}\right)=\mathrm{0} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{x}}^{\mathrm{2}} \left(\mathrm{100}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{275}\boldsymbol{\mathrm{x}}−\mathrm{144}\right)+\mathrm{30}\left(\mathrm{15}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{13}\boldsymbol{\mathrm{x}}−\mathrm{80}\right) \\ $$$$+\mathrm{225}\left(\mathrm{9}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{30}\boldsymbol{\mathrm{x}}+\mathrm{25}\right)−\mathrm{125}\left(\mathrm{60}\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{127}\boldsymbol{\mathrm{x}}+\mathrm{45}\right) \\ $$$$ \\ $$$$\mathrm{100}\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{275}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{5269}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{9515}\boldsymbol{\mathrm{x}}−\mathrm{2400}=\mathrm{0} \\ $$$$\:\: \\ $$$$\:\mathrm{Resolution}\:\mathrm{d}\:\mathrm{equation}\:\mathrm{donne} \\ $$$$ \\ $$$$\:\mathrm{valeur}\:\mathrm{accepte}\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{4},\mathrm{513}\:\boldsymbol{\mathrm{cm}} \\ $$$$\:\:\: \\ $$$$ \\ $$
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
$${NP}={PC}=\mathrm{3} \\ $$$${MQ}=\frac{{NP}}{\mathrm{2}}=\mathrm{1}.\mathrm{5} \\ $$$${QP}=\frac{{AP}}{\mathrm{2}}=\mathrm{3}.\mathrm{5} \\ $$$${BQ}={QC} \\ $$$${x}+\mathrm{1}.\mathrm{5}=\mathrm{3}.\mathrm{5}+\mathrm{3} \\ $$$$\Rightarrow{x}=\mathrm{5}\:\checkmark \\ $$