Răspuns:
Explicație pas cu pas:
[tex]tgA=\dfrac{1}{3},~~~\dfrac{sinA}{cosA}=\dfrac{1}{3},~deci~cosA=3sinA,~~dar~~sin^{2}A+cos^{2}A =1,~~sin^{2}A+(3sinA)^{2}=1,~~10sin^{2}A =1~~~sin^{2}A=\dfrac{1}{10},~~~sinA=\dfrac{\sqrt{10} }{10}\\Atunci~cosA=\dfrac{3\sqrt{10} }{10}. ~~A+B=90^{o},~deci~A=90^{o}-B.\\cosA=cos(90^{o}-B)=sinB=\dfrac{3\sqrt{10} }{10}.[/tex]