Question-43465

Question Number 43465 by ajfour last updated on 11/Sep/18

Commented by ajfour last updated on 11/Sep/18

$${Find}\:\boldsymbol{{x}},\:{and}\:\boldsymbol{{a}}. \\ $$

Answered by math1967 last updated on 11/Sep/18

(x/1)=((√(1−x^2 ))/x) [similarity of triangle] x^2 =(√(1−x^2 )) x^4 +x^2 −1=0 ⇒x^2 =((−1±(√5))/2) ⇒x=(√((((√5)−1)/2) )) a=((1+x^2 )/x)⇒(((√5)+1)/2)×(√(2/( (√5)−1)))

$$\frac{{x}}{\mathrm{1}}=\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{{x}}\:\:\left[{similarity}\:{of}\:{triangle}\right] \\ $$$${x}^{\mathrm{2}} =\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$${x}^{\mathrm{4}} +{x}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}\:\:\Rightarrow{x}=\sqrt{\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:\:} \\ $$$${a}=\frac{\mathrm{1}+{x}^{\mathrm{2}} }{{x}}\Rightarrow\frac{\sqrt{\mathrm{5}}+\mathrm{1}}{\mathrm{2}}×\sqrt{\frac{\mathrm{2}}{\:\sqrt{\mathrm{5}}−\mathrm{1}}} \\ $$

Commented by ajfour last updated on 11/Sep/18

$${short}\:{and}\:{good}\:{way},\:{Sir}\:;\:{thanks}. \\ $$

Commented by MJS last updated on 12/Sep/18

$$\mathrm{true}.\:\mathrm{simplified}\:{a}=\sqrt{\mathrm{2}+\sqrt{\mathrm{5}}} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Sep/18

x+b=a tanθ=(x/1) cosθ=(x/1) ((sinθ)/(cosθ))=cosθ t=1−t^2 t^2 +t−1=0 t=((−1±(√(1+4)))/2)=((−1±(√5))/2) t=sinθ=(((√5) −1)/2) cos^2 θ=1−(((5+1−2(√5) )/4))=((4−6+2(√5))/4)=(((√5) −1)/2) x=(√(((√5) −1)/2)) sinθ=(((√(x^2 +1)) )/a) a=(((√(x^2 +1)) )/(sinθ))=(((√((((√5) −1)/2)+1)) )/(((√5) −1)/2))=((√(((√5) +1)/2))/(((√5) −1)/2))

$${x}+{b}={a} \\ $$$${tan}\theta=\frac{{x}}{\mathrm{1}} \\ $$$${cos}\theta=\frac{{x}}{\mathrm{1}} \\ $$$$\frac{{sin}\theta}{{cos}\theta}={cos}\theta \\ $$$${t}=\mathrm{1}−{t}^{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} +{t}−\mathrm{1}=\mathrm{0} \\ $$$${t}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{4}}}{\mathrm{2}}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${t}={sin}\theta=\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$${cos}^{\mathrm{2}} \theta=\mathrm{1}−\left(\frac{\mathrm{5}+\mathrm{1}−\mathrm{2}\sqrt{\mathrm{5}}\:}{\mathrm{4}}\right)=\frac{\mathrm{4}−\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{4}}=\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}} \\ $$$${x}=\sqrt{\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}}\: \\ $$$${sin}\theta=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:}{{a}} \\ $$$${a}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:}{{sin}\theta}=\frac{\sqrt{\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}+\mathrm{1}}\:}{\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}}=\frac{\sqrt{\frac{\sqrt{\mathrm{5}}\:+\mathrm{1}}{\mathrm{2}}}}{\frac{\sqrt{\mathrm{5}}\:−\mathrm{1}}{\mathrm{2}}} \\ $$

Commented by ajfour last updated on 11/Sep/18

$${Thanks}\:{sir}\:! \\ $$

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