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Question Number 143332 by bramlexs22 last updated on 13/Jun/21

Commented by soumyasaha last updated on 13/Jun/21

$$\mathrm{x}=\sqrt{\frac{3+\sqrt{5}}{2}}$$

Commented by mr W last updated on 13/Jun/21

$$\sqrt{\frac{3+\sqrt{5}}{2}}=\sqrt{\frac{6+2\sqrt{5}}{4}}=\sqrt{\frac{{(\sqrt{5}+1)}^2}{2^2}}=\frac{\sqrt{5}+1}{2}$$

Answered by nimnim last updated on 13/Jun/21

$$cos\theta =\frac{x^2+1^2-{\left(\sqrt{\sqrt{5}}\right)}^2}{2.1.x}=\frac{x^2+1-\sqrt{5}}{2x}$$$$alsocos(180-\theta )=\frac{x^2+1^2-{\left(\sqrt{5}\right)}^2}{2.x.1}$$$$-cos\theta =\frac{x^2-4}{2x}\Rightarrow cos\theta =\frac{-x^2+4}{2x}$$$$\therefore \frac{x^2+1-\sqrt{5}}{2x}=\frac{-x^2+4}{2x}$$$$\Rightarrow 2x^2=3+\sqrt{5}\iff 4x^2=6+2\sqrt{5}$$$$\Rightarrow x^2=\frac{{(1+\sqrt{5})}^2}{4}\iff x=\pm \frac{1+\sqrt{5}}{2}$$$$\Rightarrow x=\frac{1+\sqrt{5}}{2}=\phi (aslengthx>0)$$

Commented by nimnim last updated on 13/Jun/21

Answered by mr W last updated on 13/Jun/21

Commented by mr W last updated on 13/Jun/21

$$1isamedianoftrianglewith$$$$sides\sqrt{\sqrt{5}},\sqrt{5}and2x.$$$$4\times 1^2=2{\left(\sqrt{\sqrt{5}}\right)}^2+2{\left(\sqrt{5}\right)}^2-{\left(2x\right)}^2$$$${(\sqrt{5}+1)}^2={\left(2x\right)}^2$$$$\Rightarrow x=\frac{\sqrt{5}+1}{2}$$

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