Salut,
sin²A + cos²A = 1, deci cos²A = 1 -- sin²A.
[tex]cosA=\pm\sqrt{1-sin^2A}=\pm\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\pm\sqrt{1-\dfrac{25}{169}}=\\\\=\pm\sqrt{\dfrac{144}{149}}=\pm\dfrac{12}{13}.\ Cum\ A\in(0,\ 90^{\circ}),\ deci\ cosA>0\Rightarrow cosA=+\dfrac{12}{13}.[/tex]
Green eyes.
