Question-210473

Question Number 210473 by Ismoiljon_008 last updated on 10/Aug/24

Answered by mrdiane last updated on 11/Aug/24

on a { ((7^2 =x^2 +5^2 −10xcos(θ))),((7^2 =x^2 +(x+5)^2 −2x(x+5)cos(Π−θ))) :} ⇔ { ((7^2 =x^2 +5^2 −10xcos(θ) (1))),((7^2 =x^2 +(x+5)^2 +2x(x+5)cos(θ) (2))) :} (1)=(2) on a 2(x+5)cos(θ)+x+10=−10cos(θ) ⇔cos(θ)=((−x−10)/(2x+20)) =((−1)/2) pour car x≥0 ⇔x^2 −24= dans (1) on a x^2 +5x−24=0 ⇔x=3 ou x=−8 comme x≥0 donc x=3 x^2 +5x−24=0

$${on}\:{a}\:\begin{cases}{\mathrm{7}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{10}{xcos}\left(\theta\right)}\\{\mathrm{7}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left({x}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{2}{x}\left({x}+\mathrm{5}\right){cos}\left(\Pi−\theta\right)}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\mathrm{7}^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{10}{xcos}\left(\theta\right)\:\left(\mathrm{1}\right)}\\{\mathrm{7}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left({x}+\mathrm{5}\right)^{\mathrm{2}} +\mathrm{2}{x}\left({x}+\mathrm{5}\right){cos}\left(\theta\right)\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)=\left(\mathrm{2}\right)\:{on}\:{a}\:\mathrm{2}\left({x}+\mathrm{5}\right){cos}\left(\theta\right)+{x}+\mathrm{10}=−\mathrm{10}{cos}\left(\theta\right) \\ $$$$\Leftrightarrow{cos}\left(\theta\right)=\frac{−{x}−\mathrm{10}}{\mathrm{2}{x}+\mathrm{20}}\:=\frac{−\mathrm{1}}{\mathrm{2}}\:{pour}\:{car}\:{x}\geqslant\mathrm{0} \\ $$$$\Leftrightarrow{x}^{\mathrm{2}} −\mathrm{24}= \\ $$$${dans}\:\left(\mathrm{1}\right)\:{on}\:{a}\:{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{24}=\mathrm{0}\:\Leftrightarrow{x}=\mathrm{3}\:{ou}\:{x}=−\mathrm{8}\:{comme}\:{x}\geqslant\mathrm{0} \\ $$$${donc}\:{x}=\mathrm{3} \\ $$$${x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{24}=\mathrm{0} \\ $$$$ \\ $$

Answered by efronzo1 last updated on 11/Aug/24

cos θ = ((x^2 +(x+5)^2 −49)/(2x(x+5))) cos (180°−θ)= ((25+x^2 −49)/(2.5.x)) ⇒−cos θ = ((x^2 −24)/(10x)) ⇒− ((2x^2 +10x−24)/(2x(x+5))) = ((x^2 −24)/(10x)) ⇒−(10x^2 +50x−120) = (x^2 −24)(x+5) ⇒−10x^2 −50x+120=x^3 +5x^2 −24x−120 ⇒x^3 +15x^2 +26x−240=0 ⇒(x−3)(x+8)(x+10)=0 ∴ x = 3

$$\:\:\mathrm{cos}\:\theta\:=\:\frac{\mathrm{x}^{\mathrm{2}} +\left(\mathrm{x}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{49}}{\mathrm{2x}\left(\mathrm{x}+\mathrm{5}\right)} \\ $$$$\:\mathrm{cos}\:\left(\mathrm{180}°−\theta\right)=\:\frac{\mathrm{25}+\mathrm{x}^{\mathrm{2}} −\mathrm{49}}{\mathrm{2}.\mathrm{5}.\mathrm{x}} \\ $$$$\:\Rightarrow−\mathrm{cos}\:\theta\:=\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{24}}{\mathrm{10x}}\: \\ $$$$\:\Rightarrow−\:\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{10x}−\mathrm{24}}{\mathrm{2x}\left(\mathrm{x}+\mathrm{5}\right)}\:=\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{24}}{\mathrm{10x}} \\ $$$$\:\Rightarrow−\left(\mathrm{10x}^{\mathrm{2}} +\mathrm{50x}−\mathrm{120}\right)\:=\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{24}\right)\left(\mathrm{x}+\mathrm{5}\right) \\ $$$$\:\Rightarrow−\mathrm{10x}^{\mathrm{2}} −\mathrm{50x}+\mathrm{120}=\mathrm{x}^{\mathrm{3}} +\mathrm{5x}^{\mathrm{2}} −\mathrm{24x}−\mathrm{120} \\ $$$$\:\Rightarrow\mathrm{x}^{\mathrm{3}} +\mathrm{15x}^{\mathrm{2}} +\mathrm{26x}−\mathrm{240}=\mathrm{0} \\ $$$$\:\Rightarrow\left(\mathrm{x}−\mathrm{3}\right)\left(\mathrm{x}+\mathrm{8}\right)\left(\mathrm{x}+\mathrm{10}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\mathrm{x}\:=\:\mathrm{3}\: \\ $$

Answered by mr W last updated on 10/Aug/24

Commented by mr W last updated on 10/Aug/24

Method I x(y^2 +7^2 )=2x(5^2 +x^2 ) ⇒y^2 =2x^2 +1 5×7^2 +x×x^2 =(5+x)(y^2 +5x) ⇒245+x^3 =(5+x)(2x^2 +1+5x) ⇒x^3 +15x^2 +26x−240=0 ⇒(x−3)(x^2 +18x+80)=0 ⇒x=3

$$\underline{\boldsymbol{{Method}}\:\boldsymbol{{I}}} \\ $$$${x}\left({y}^{\mathrm{2}} +\mathrm{7}^{\mathrm{2}} \right)=\mathrm{2}{x}\left(\mathrm{5}^{\mathrm{2}} +{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{y}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{5}×\mathrm{7}^{\mathrm{2}} +{x}×{x}^{\mathrm{2}} =\left(\mathrm{5}+{x}\right)\left({y}^{\mathrm{2}} +\mathrm{5}{x}\right) \\ $$$$\Rightarrow\mathrm{245}+{x}^{\mathrm{3}} =\left(\mathrm{5}+{x}\right)\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}+\mathrm{5}{x}\right) \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} +\mathrm{26}{x}−\mathrm{240}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−\mathrm{3}\right)\left({x}^{\mathrm{2}} +\mathrm{18}{x}+\mathrm{80}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$

Answered by mr W last updated on 10/Aug/24

Commented by mr W last updated on 10/Aug/24

Method II AE=x+2×5 CE=((AE)/2)=(x/2)+5 CF=(x/2)+5−5=(x/2) DC^2 =7^2 −((x/2)+5)^2 =x^2 −((x/2))^2 x^2 +5x−24=0 (x−3)(x+8)=0 ⇒x=3

$$\underline{\boldsymbol{{Method}}\:\boldsymbol{{II}}} \\ $$$${AE}={x}+\mathrm{2}×\mathrm{5} \\ $$$${CE}=\frac{{AE}}{\mathrm{2}}=\frac{{x}}{\mathrm{2}}+\mathrm{5} \\ $$$${CF}=\frac{{x}}{\mathrm{2}}+\mathrm{5}−\mathrm{5}=\frac{{x}}{\mathrm{2}} \\ $$$${DC}^{\mathrm{2}} =\mathrm{7}^{\mathrm{2}} −\left(\frac{{x}}{\mathrm{2}}+\mathrm{5}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} −\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{24}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{8}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3} \\ $$

Commented by Ismoiljon_008 last updated on 10/Aug/24