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Question Number 210120 by emilagazade last updated on 31/Jul/24

Answered by mr W last updated on 31/Jul/24

Commented by mr W last updated on 31/Jul/24

cos θ=((5^2 +(8−x)^2 −(3+x)^2 )/(2×5×(8−x))) cos (180−θ)=((3^2 +(8−x)^2 −(5+x)^2 )/(2×3×(8−x)))=−cos θ ((3^2 +(8−x)^2 −(5+x)^2 )/(2×3×(8−x)))=−((5^2 +(8−x)^2 −(3+x)^2 )/(2×5×(8−x))) ⇒x=((120)/(49))

$$\mathrm{cos}\:\theta=\frac{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{8}−{x}\right)^{\mathrm{2}} −\left(\mathrm{3}+{x}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}×\left(\mathrm{8}−{x}\right)} \\ $$$$\mathrm{cos}\:\left(\mathrm{180}−\theta\right)=\frac{\mathrm{3}^{\mathrm{2}} +\left(\mathrm{8}−{x}\right)^{\mathrm{2}} −\left(\mathrm{5}+{x}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\left(\mathrm{8}−{x}\right)}=−\mathrm{cos}\:\theta \\ $$$$\frac{\mathrm{3}^{\mathrm{2}} +\left(\mathrm{8}−{x}\right)^{\mathrm{2}} −\left(\mathrm{5}+{x}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{3}×\left(\mathrm{8}−{x}\right)}=−\frac{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{8}−{x}\right)^{\mathrm{2}} −\left(\mathrm{3}+{x}\right)^{\mathrm{2}} }{\mathrm{2}×\mathrm{5}×\left(\mathrm{8}−{x}\right)} \\ $$$$\Rightarrow{x}=\frac{\mathrm{120}}{\mathrm{49}} \\ $$

Answered by mr W last updated on 31/Jul/24

an other method: ((1/5)+(1/3)+(1/x)−(1/8))^2 =2((1/5^2 )+(1/3^2 )+(1/x^2 )+(1/8^2 )) ⇒(1/x^2 )−((49)/(60x))+((2401)/(14400))=0 ⇒(1/x)=((49)/(120)) ⇒x=((120)/(49))

$${an}\:{other}\:{method}: \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{\mathrm{8}}\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{8}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{49}}{\mathrm{60}{x}}+\frac{\mathrm{2401}}{\mathrm{14400}}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\frac{\mathrm{49}}{\mathrm{120}}\:\Rightarrow{x}=\frac{\mathrm{120}}{\mathrm{49}} \\ $$

Commented by emilagazade last updated on 31/Jul/24

thank you a lot Sir

$${thank}\:{you}\:{a}\:{lot}\:{Sir} \\ $$

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