[tex]\it m(\hat{A}) = \pi-\Big(\dfrac{\pi}{8}+\dfrac{3\pi}{8}\Big)= \pi-\dfrac{4\pi}{8}=\pi-\dfrac{\pi}{2}=\dfrac{\pi}{2}[/tex]
Deci, triunghiul ABC este dreptunghic în A.
[tex]\it cos^2B =\Big(\dfrac{c}{a}\Big)^2 \Rightarrow cos^2\dfrac{\pi}{8} =\Big(\dfrac{c}{a}\Big)^2\ \ \ \ (1)\\ \\ \\ cos2\cdot\dfrac{\pi}{8} =2cos^2\dfrac{\pi}{8}-1 \Rightarrow cos\dfrac{\pi}{4}=2cos^2\dfrac{\pi}{8}-1 \Rightarrow cos^2\dfrac{\pi}{8} =\dfrac{cos\dfrac{\pi}{4}+1}{2} \Rightarrow\\ \\ \\ \Rightarrow cos^2\dfrac{\pi}{8}=\dfrac{\dfrac{\sqrt2}{2}+1}{2} \Rightarrow cos^2\dfrac{\pi}{8}= \dfrac{\sqrt2+2}{4}\ \ \ \ (2)[/tex]
[tex]\it (1),\ (2) \Rightarrow 4\Big(\dfrac{c}{a}\Big)^2 =4cos^2\dfrac{\pi}{8}=4\cdot\dfrac{\sqrt2+2}{4}=\sqrt2+2\ \ \ \ \ (3)\\ \\ \\ sin^2\dfrac{\pi}{8}=1-cos^2\dfrac{\pi}{2} \stackrel{(2)}{\Longrightarrow}\ sin^2\dfrac{\pi}{8}=1-\dfrac{\sqrt2+2}{4} \Rightarrow sin^2\dfrac{\pi}{8}=\dfrac{2-\sqrt2}{4}[/tex]
[tex]\it tg^2\dfrac{\pi}{8}=\Big(\dfrac{b}{c}\Big)^2 \Rightarrow \dfrac{sin^2\dfrac{\pi}{8}}{cos^2\dfrac{\pi}{8}}=\Big(\dfrac{b}{c}\Big)^2 \Rightarrow \dfrac{\dfrac{2-\sqrt2}{4}}{\dfrac{2+\sqrt2}{4}}=\Big(\dfrac{b}{c}\Big)^2 \Rightarrow\\ \\ \\ \Rightarrow \Big(\dfrac{b}{c}\Big)^2=\dfrac{^{2-\sqrt2)}2-\sqrt2}{\ 2+\sqrt2}=\dfrac{(2-\sqrt2)^2}{2} =\Big(\dfrac{2-\sqrt2}{\sqrt2}\Big)^2\Rightarrow \\ \\ \\ \Rightarrow \dfrac{b}{c}=\dfrac{2-\sqrt2}{\sqrt2} =\dfrac{\sqrt2(\sqrt2-1)}{\sqrt2}=\sqrt2-1\ \ \ (4)[/tex]
Folosind relațiile (3), (4), expresia din enunț devine:
[tex]\it2+\sqrt2-\sqrt2+1=3[/tex]
