Odpowiedź :
[tex]sin30' + sin45' + sin60' = \dfrac{1}{2}+\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{3}}{2} = \dfrac{1+\sqrt{2}+\sqrt{3}}{2}\\\\tg30'+tg45'+tg60'=\dfrac{\sqrt{3}}{3}+1+\sqrt{3} = \dfrac{\sqrt{3}}{3}+1+\dfrac{3\sqrt{3}}{3} = \dfrac{4\sqrt{3}}{3}+1\\\\sin30'+cos30'+tg45'= \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}+1=\dfrac{\sqrt{3}}{2}+\dfrac{3}{2} = \dfrac{3+\sqrt{3}}{2}[/tex]