Răspuns :
[tex]\displaystyle Viteza \ de \ deplasare \ a \ corpului \ este \ constanta \ , \ deci \ este \ in \\ \\ echilibru \ mecanic \ . \ Conditiile \ de \ echilibru \ sunt : \\ \\ \vec{N}+\vec{G_n}=0\implies N=G_n \\ \\ \vec{F}+\vec{G_t}+\vec{F_f}=0\implies F=G_t+F_f \\ \\ Stim \ ca \ F=mg: \\ \\ mg=G\sin\alpha+N\mu \\ \\ mg=mg\sin\alpha+mg\mu\cos\alpha \\ \\ mg=mg(\sin\alpha+\mu\cos\alpha) \\ \\ \sin\alpha+\mu\cos\alpha=0 \\ \\ Conform \ teoremei \ fundamentale \ a \ trigonometriei \ : \ \sin^2\alpha+\cos^2\alpha=1\implies\sin\alpha=\sqrt{1-\cos^2\alpha} \\ \\ Substituim: \\ \\ \sqrt{1-\cos^2\alpha}+\mu\cos\alpha=0 \\ \\ \sqrt{1-\cos^2\alpha}=-\mu\cos\alpha|^2 \\ \\ 1-\cos^2\alpha=\mu^2\cos^2\alpha \\ \\ \mu^2\cos^2\alpha+\cos^2\alpha=1 \\ \\ 4\cos^2\alpha=1 \\ \\ \cos^2\alpha=\dfrac{1}{4} \\ \\ \cos\alpha=\sqrt{\dfrac{1}{4}} \\ \\ \cos\alpha=\dfrac{1}{2}\implies\alpha=60\textdegree[/tex]