Let us consider a 30°-60°-90°△ABC right angled at B in which ∠C = 30⁰ and ∠A = 60⁰ with hypotenuse AC = 10√2 units.
In △ABC,
[tex]\longrightarrow[/tex]sin ∠C = [tex]\sf \dfrac{Perpendicular}{Hypotenuse}[/tex]
[tex]\longrightarrow[/tex]sin ∠C = [tex]\sf \dfrac{AB}{AC}[/tex]
[tex]\longrightarrow[/tex]sin 30⁰ = [tex]\sf \dfrac{AB}{10\sqrt{2}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf \dfrac{1}{2}= \dfrac{AB}{10\sqrt{2}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf AB = \dfrac{10\sqrt{2}}{2}[/tex]
[tex]\longrightarrow[/tex][tex]\sf AB = 5\sqrt{2}\:units[/tex]
[tex]\\[/tex]
In △ABC,
[tex]\longrightarrow[/tex]cos ∠C = [tex]\sf \dfrac{Base}{Hypotenuse}[/tex]
[tex]\longrightarrow[/tex]cos ∠C = [tex]\sf \dfrac{BC}{AC}[/tex]
[tex]\longrightarrow[/tex]cos 30⁰ = [tex]\sf \dfrac{BC}{10\sqrt{2}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf \dfrac{\sqrt{3}}{2}= \dfrac{BC}{10\sqrt{2}}[/tex]
[tex]\longrightarrow[/tex][tex]\sf BC = \dfrac{10\sqrt{2}\times \sqrt{3}}{2}[/tex]
[tex]\longrightarrow[/tex][tex]\sf BC = \dfrac{10\sqrt{2\times 3}}{2}[/tex]
[tex]\longrightarrow[/tex][tex]\sf BC = \dfrac{10\sqrt{6}}{2}[/tex]
[tex]\longrightarrow[/tex][tex]\sf BC = 5\sqrt{6}\: units[/tex]
Thus , the length of other two sides of triangles are 5√2 and 5√6 units.