Question Number 212783 by Spillover last updated on 23/Oct/24
Commented by York12 last updated on 25/Oct/24
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Answered by A5T last updated on 24/Oct/24
Commented by A5T last updated on 24/Oct/24
![a=(√(s^2 +(s^2 /4)))−(s/2)=((s(√5))/2)−(s/2); cosθ=((s/2)/((s(√5))/2))=((√5)/5) s^2 =((s/2))^2 +((s/2)+b)^2 +s×((s+2b)/2)×((√5)/5) ⇒b^2 +(1+((√5)/5))bs+s^2 ((((√5)−5)/(10)))=0 ⇒b=((−(1+((√5)/5))s+s(√((1+ ((√5)/5))^2 −4((((√5)−5)/(10))))))/2) =((s(−1−((√5)/5)+((4(√5))/( 5))))/2)=((s(−1+((3(√5))/5)))/2) [since b>0] ⇒(a/b)=(((√5)−1)/(−1+((3(√5))/5)))=ϕ^2 +1](https://www.tinkutara.com/question/Q212792.png)
$${a}=\sqrt{{s}^{\mathrm{2}} +\frac{{s}^{\mathrm{2}} }{\mathrm{4}}}−\frac{{s}}{\mathrm{2}}=\frac{{s}\sqrt{\mathrm{5}}}{\mathrm{2}}−\frac{{s}}{\mathrm{2}};\:\:{cos}\theta=\frac{\frac{{s}}{\mathrm{2}}}{\frac{{s}\sqrt{\mathrm{5}}}{\mathrm{2}}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$${s}^{\mathrm{2}} =\left(\frac{{s}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{s}}{\mathrm{2}}+{b}\right)^{\mathrm{2}} +{s}×\frac{{s}+\mathrm{2}{b}}{\mathrm{2}}×\frac{\sqrt{\mathrm{5}}}{\mathrm{5}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} +\left(\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\right){bs}+{s}^{\mathrm{2}} \left(\frac{\sqrt{\mathrm{5}}−\mathrm{5}}{\mathrm{10}}\right)=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{−\left(\mathrm{1}+\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\right){s}+{s}\sqrt{\left(\mathrm{1}+\:\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{4}\left(\frac{\sqrt{\mathrm{5}}−\mathrm{5}}{\mathrm{10}}\right)}}{\mathrm{2}} \\ $$$$=\frac{{s}\left(−\mathrm{1}−\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}+\frac{\mathrm{4}\sqrt{\mathrm{5}}}{\:\mathrm{5}}\right)}{\mathrm{2}}=\frac{{s}\left(−\mathrm{1}+\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{5}}\right)}{\mathrm{2}}\:\:\:\:\left[{since}\:{b}>\mathrm{0}\right] \\ $$$$\Rightarrow\frac{{a}}{{b}}=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{−\mathrm{1}+\frac{\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{5}}}=\varphi^{\mathrm{2}} +\mathrm{1} \\ $$
Commented by Spillover last updated on 24/Oct/24
$${thank}\:{you} \\ $$