Założenia:

$\alpha +\beta ={90}^{\circ }$  

$\alpha ,\beta \in \left(0^{\circ },{90}^{\circ }\right)$  

$\sin \alpha -\sin \beta =\frac{1}{2}$     

 

 

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$\sin \beta =\sin \left({90}^{\circ }-\alpha \right)=\cos \alpha$    

 

 

 

$\sin \alpha -\cos \alpha =\frac{1}{2}$     

$\sin \alpha =\cos \alpha +\frac{1}{2}$  

 

Skorzystajmy z jedynki trygonometrycznej:   

${\sin }^2\alpha +{\cos }^2\alpha =1$  

${\left({\cos }^2\alpha +\frac{1}{2}\right)}^2+{\cos }^2\alpha =1$  

${\cos }^2\alpha +\cos \alpha +\frac{1}{4}+{\cos }^2\alpha =1$  

$2{\cos }^2\alpha +\cos \alpha -\frac{3}{4}=0$  

 

$\Delta =1^2-4\cdot 2\cdot \left(-\frac{3}{4}\right)=1+6=7$  

 

${\cos }_1\alpha =\frac{-1-\sqrt{7}}{2\cdot 2}<0$   

 

${\cos }_2\alpha =\frac{-1+\sqrt{7}}{2\cdot 2}=\frac{-1+\sqrt{7}}{4}$  

 

$\sin \alpha =\frac{-1+\sqrt{7}}{4}+\frac{1}{2}=\frac{-1+\sqrt{7}}{4}+\frac{2}{4}=\frac{1+\sqrt{7}}{4}$    

 

$\mathrm{tg}\alpha =\frac{\sin \alpha }{\cos \alpha }=\frac{\frac{1+\sqrt{7}}{4}}{\frac{-1+\sqrt{7}}{4}}=\frac{1+\sqrt{7}}{\sqrt{7}-1}=\frac{1+\sqrt{7}}{\sqrt{7}-1}\cdot \frac{\sqrt{7}+1}{\sqrt{7}+1}=\frac{{\left(1+\sqrt{7}\right)}^2}{7-1}=\frac{1+2\sqrt{7}+7}{6}=\frac{8+2\sqrt{7}}{6}=\frac{4+\sqrt{7}}{3}$   

 

$\mathrm{ctg}\alpha =\frac{3}{4+\sqrt{7}}=\frac{3}{4+\sqrt{7}}\cdot \frac{4-\sqrt{7}}{4-\sqrt{7}}=\frac{3\left(4-\sqrt{7}\right)}{16-7}=\frac{12-3\sqrt{7}}{9}=\frac{4-\sqrt{7}}{3}$  

 

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$3a^2+3b^2=8ab\mid :ab$  

 

$3\frac{a}{b}+3\frac{b}{a}=8$  

 

$\frac{a}{b}=\mathrm{tg}\alpha$  

$\frac{b}{a}=\mathrm{ctg}\alpha$  

 

$L=3\frac{a}{b}+3\frac{b}{a}=3\cdot \frac{4+\sqrt{7}}{3}+3\cdot \frac{4-\sqrt{7}}{3}=4+\sqrt{7}+4-\sqrt{7}=8=P$   - a więc długości a, b spełniają warunek z treści zadania.