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

Commented by ajfour last updated on 11/Sep/18

$$Find\mathit{x},and\mathit{a}.$$

Answered by math1967 last updated on 11/Sep/18

$$\frac{x}{1}=\frac{\sqrt{1-x^2}}{x}\left[similarityoftriangle\right]$$$$x^2=\sqrt{1-x^2}$$$$x^4+x^2-1=0$$$$\Rightarrow x^2=\frac{-1\pm \sqrt{5}}{2}\Rightarrow x=\sqrt{\frac{\sqrt{5}-1}{2}}$$$$a=\frac{1+x^2}{x}\Rightarrow \frac{\sqrt{5}+1}{2}\times \sqrt{\frac{2}{\sqrt{5}-1}}$$

Commented by ajfour last updated on 11/Sep/18

$$shortandgoodway,Sir;thanks.$$

Commented by MJS last updated on 12/Sep/18

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

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

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

Commented by ajfour last updated on 11/Sep/18

$$Thankssir!$$

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