Question-187608

Question Number 187608 by ajfour last updated on 19/Feb/23

Answered by a.lgnaoui last updated on 19/Feb/23

tan α=(X/c)=((X+1)/(2X+c))⇒ (2X+c)X=c(X+1) 2X^2 +cX=cX+c 2X^2 =c X=(√(c/2))

$$\mathrm{tan}\:\alpha=\frac{\mathrm{X}}{{c}}=\frac{\mathrm{X}+\mathrm{1}}{\mathrm{2X}+{c}}\Rightarrow \\ $$$$\left(\mathrm{2X}+{c}\right)\mathrm{X}={c}\left(\mathrm{X}+\mathrm{1}\right) \\ $$$$\mathrm{2X}^{\mathrm{2}} +{c}\mathrm{X}={c}\mathrm{X}+{c} \\ $$$$\:\:\:\:\mathrm{2X}^{\mathrm{2}} ={c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{X}=\sqrt{\frac{{c}}{\mathrm{2}}} \\ $$

Commented by a.lgnaoui last updated on 19/Feb/23

Commented by ajfour last updated on 19/Feb/23

how verical is 2x ?

Commented by a.lgnaoui last updated on 19/Feb/23

$${OH}\:{with}\:{OH}+{C}=\mathrm{2X}+\mathrm{C} \\ $$$$ \\ $$

Commented by a.lgnaoui last updated on 19/Feb/23

Commented by a.lgnaoui last updated on 19/Feb/23

$${if}\:{the}\:{quadrilatere}\:{is}\:{not} \\ $$$${squart}\:{that}\:{is}\:{an}\:{authere} \\ $$$${value}\: \\ $$$${your}\:{solution}\:{is}\:{general} \\ $$$${thanks}. \\ $$

Commented by mr W last updated on 19/Feb/23

$${you}\:{just}\:{ignore}\:{the}\:{semicircle}\:{in}\: \\ $$$${order}\:{to}\:{be}\:{able}\:{to}\:{solve}\:{the}\:{problem}, \\ $$$${but}\:{this}\:{is}\:{not}\:{a}\:{real}\:{solution},\:{because} \\ $$$${you}\:{have}\:{changed}\:{the}\:{question}. \\ $$

Commented by mr W last updated on 19/Feb/23

Commented by a.lgnaoui last updated on 19/Feb/23

$${h}\:{can}\:{take}\:{value}\:\mathrm{2}{x}. \\ $$

Commented by mr W last updated on 20/Feb/23

$${it}'{s}\:{then}\:{not}\:{the}\:{asked}\:{question}, \\ $$$${therefore}\:{non}−{sense}. \\ $$

Commented by ajfour last updated on 20/Feb/23

yeah if the company agrees it can give me the car for 5$.

Answered by ajfour last updated on 19/Feb/23

tan α=(x/c)=(1/h)=(h/(2x+1)) ⇒ h^2 =2x+1=(c^2 /x^2 ) ⇒ 2x^3 +x^2 =c^2 or x(√(2x+1))=c

$$\mathrm{tan}\:\alpha=\frac{{x}}{{c}}=\frac{\mathrm{1}}{{h}}=\frac{{h}}{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\Rightarrow\:\:{h}^{\mathrm{2}} =\mathrm{2}{x}+\mathrm{1}=\frac{{c}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\mathrm{2}{x}^{\mathrm{3}} +{x}^{\mathrm{2}} ={c}^{\mathrm{2}} \\ $$$${or}\:\:\:\:{x}\sqrt{\mathrm{2}{x}+\mathrm{1}}={c} \\ $$$$\:\:\: \\ $$

Answered by mr W last updated on 19/Feb/23

h^2 =1×(2x+1) ⇒h=(√(2x+1)) (c/x)=(h/1) ⇒x(√(2x+1))=c x^2 (2x+1)−c^2 =0 (1/x^3 )−(1/(c^2 x))−(2/c^2 )=0 ⇒(1/x)=(((1/c^2 )((√(1−(1/(27c^2 ))))+1)))^(1/3) −(((1/c^2 )((√(1−(1/(27c^2 ))))−1)))^(1/3) ⇒x=(1/( (((1/c^2 )((√(1−(1/(27c^2 ))))+1)))^(1/3) −(((1/c^2 )((√(1−(1/(27c^2 ))))−1)))^(1/3) ))

$${h}^{\mathrm{2}} =\mathrm{1}×\left(\mathrm{2}{x}+\mathrm{1}\right)\:\Rightarrow{h}=\sqrt{\mathrm{2}{x}+\mathrm{1}} \\ $$$$\frac{{c}}{{x}}=\frac{{h}}{\mathrm{1}}\: \\ $$$$\Rightarrow{x}\sqrt{\mathrm{2}{x}+\mathrm{1}}={c} \\ $$$${x}^{\mathrm{2}} \left(\mathrm{2}{x}+\mathrm{1}\right)−{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{{c}^{\mathrm{2}} {x}}−\frac{\mathrm{2}}{{c}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{x}}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}+\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}−\mathrm{1}\right)} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}+\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{{c}^{\mathrm{2}} }\left(\sqrt{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{27}{c}^{\mathrm{2}} }}−\mathrm{1}\right)}} \\ $$

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