Question Number 202614 by a.lgnaoui last updated on 30/Dec/23
$$\mathrm{find}\:\mathrm{x} \\ $$$$ \\ $$
Commented by a.lgnaoui last updated on 30/Dec/23
Commented by a.lgnaoui last updated on 31/Dec/23
$$\left(\mathrm{x}+\mathrm{5}\right)\mathrm{sin}\:\boldsymbol{\varphi}=\mathrm{7sin}\:\boldsymbol{\lambda}+\mathrm{xsin}\:\boldsymbol{\varphi}\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\mathrm{5sin}\:\boldsymbol{\varphi}\:\:\:\:\:\:\:\:\:\:=\mathrm{7sin}\:\boldsymbol{\lambda}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{AB}}^{\mathrm{2}} =\boldsymbol{\mathrm{AC}}^{\mathrm{2}} +\boldsymbol{\mathrm{BC}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{AC}}.\boldsymbol{\mathrm{BC}}.\mathrm{cos}\:\boldsymbol{\varphi} \\ $$$$ \\ $$$$\:\:\:\mathrm{7}^{\mathrm{2}} =\:\:\left(\boldsymbol{\mathrm{x}}+\mathrm{5cos}\:\boldsymbol{\varphi}\right)^{\mathrm{2}} +\left(\mathrm{5sin}\:\boldsymbol{\varphi}\right)^{\mathrm{2}} \\ $$$$\:\:\mathrm{49}=\mathrm{x}^{\mathrm{2}} +\mathrm{10}\boldsymbol{\mathrm{x}}\mathrm{cos}\:\boldsymbol{\varphi}+\mathrm{25} \\ $$$$\:\:\mathrm{24}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{10}\boldsymbol{\mathrm{x}}\mathrm{cos}\:\boldsymbol{\varphi}\:\:\:;\:\mathrm{cos}\:\boldsymbol{\varphi}=\frac{\mathrm{24}−\boldsymbol{\mathrm{x}}^{\mathrm{2}} }{\mathrm{10}\boldsymbol{\mathrm{x}}} \\ $$$$\:\:\mathrm{49}=\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\left(\boldsymbol{\mathrm{x}}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{2}\boldsymbol{\mathrm{x}}\left(\boldsymbol{\mathrm{x}}+\mathrm{5}\right)\mathrm{cos}\:\boldsymbol{\varphi} \\ $$$$ \\ $$$$\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{3}} +\mathrm{15}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{26}\boldsymbol{\mathrm{x}}−\mathrm{240}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}=\mathrm{3}\: \\ $$$$\:\:\: \\ $$
Commented by a.lgnaoui last updated on 31/Dec/23
Answered by mr W last updated on 31/Dec/23
$$\mathrm{cos}\:\alpha=−\mathrm{cos}\:\beta \\ $$$$\frac{{x}^{\mathrm{2}} +\left({x}+\mathrm{5}\right)^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}{x}\left({x}+\mathrm{5}\right)}=−\frac{{x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}{x}} \\ $$$${x}^{\mathrm{3}} +\mathrm{15}{x}^{\mathrm{2}} +\mathrm{26}{x}−\mathrm{240}=\mathrm{0} \\ $$$$\left({x}−\mathrm{3}\right)\left({x}+\mathrm{8}\right)\left({x}+\mathrm{10}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{3}\:\checkmark \\ $$
Commented by mr W last updated on 31/Dec/23
Answered by AST last updated on 30/Dec/23
![7^2 =5^2 +x^2 −10xcos(θ)⇒24=x^2 −10xcos(θ) 7^2 =x^2 +(5+x)^2 −2x(5+x)cos(180−θ) ⇒49=x^2 +25+10x+x^2 +10xcos(θ)+2x^2 cos(θ) ⇒24=2x^2 +10x+10xcos(θ)+2x^2 cos(θ) ⇒x+10+20cos(θ)+2xcos(θ)=0 x(1+2cos(θ))=−10(1+2cos(θ) x=−10 or cos(θ)=((−1)/2) ⇒x^2 +5x−24=0⇒x=3 [θ is the angle between length (5+x) and (x)]](https://www.tinkutara.com/question/Q202621.png)
$$\mathrm{7}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{10}{xcos}\left(\theta\right)\Rightarrow\mathrm{24}={x}^{\mathrm{2}} −\mathrm{10}{xcos}\left(\theta\right) \\ $$$$\mathrm{7}^{\mathrm{2}} ={x}^{\mathrm{2}} +\left(\mathrm{5}+{x}\right)^{\mathrm{2}} −\mathrm{2}{x}\left(\mathrm{5}+{x}\right){cos}\left(\mathrm{180}−\theta\right) \\ $$$$\Rightarrow\mathrm{49}={x}^{\mathrm{2}} +\mathrm{25}+\mathrm{10}{x}+{x}^{\mathrm{2}} +\mathrm{10}{xcos}\left(\theta\right)+\mathrm{2}{x}^{\mathrm{2}} {cos}\left(\theta\right) \\ $$$$\Rightarrow\mathrm{24}=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{10}{xcos}\left(\theta\right)+\mathrm{2}{x}^{\mathrm{2}} {cos}\left(\theta\right) \\ $$$$\Rightarrow{x}+\mathrm{10}+\mathrm{20}{cos}\left(\theta\right)+\mathrm{2}{xcos}\left(\theta\right)=\mathrm{0} \\ $$$${x}\left(\mathrm{1}+\mathrm{2}{cos}\left(\theta\right)\right)=−\mathrm{10}\left(\mathrm{1}+\mathrm{2}{cos}\left(\theta\right)\right. \\ $$$${x}=−\mathrm{10}\:{or}\:{cos}\left(\theta\right)=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +\mathrm{5}{x}−\mathrm{24}=\mathrm{0}\Rightarrow{x}=\mathrm{3} \\ $$$$\left[\theta\:{is}\:{the}\:{angle}\:{between}\:{length}\:\left(\mathrm{5}+{x}\right)\:{and}\:\left({x}\right)\right] \\ $$