The Pythagorean theorem states:
[tex]c^2=a^2+b^2[/tex]where a and b are the legs and c is the hypotenuse of a right triangle.
Substituting with b = 5, and c = 5√3, and solving for a (opposite side to angle A):
[tex]\begin{gathered} (5\sqrt[]{3})^2=a^2+5^2 \\ 5^2(\sqrt[]{3})^2=a^2+25 \\ 25\cdot3=a^2+25 \\ 75-25=a^2 \\ \sqrt[]{50}=a \\ \sqrt[]{25}\cdot\sqrt[]{2}=a \\ 5\sqrt[]{2}=a \end{gathered}[/tex]By definition:
[tex]\sin (angle)=\frac{\text{opposite}}{\text{hypotenuse}}[/tex]Considering angle A, the opposite side is 5√2 and the hypotenuse is 5√3. Substituting with this information, we get:
[tex]\begin{gathered} \sin A=\frac{5\sqrt[]{2}}{5\sqrt[]{3}} \\ \sin A=\frac{\sqrt[]{2}}{\sqrt[]{3}}\cdot\frac{\sqrt[]{3}}{\sqrt[]{3}} \\ \sin A=\frac{\sqrt[]{2\cdot3}}{(\sqrt[]{3})^2} \\ \sin A=\frac{\sqrt[]{6}}{3} \end{gathered}[/tex]Considering angle B, the opposite side is 5 and the hypotenuse is 5√3. Substituting with this information, we get:
[tex]\begin{gathered} \sin B=\frac{5}{5\sqrt[]{3}} \\ \sin B=\frac{1}{\sqrt[]{3}}\cdot\frac{\sqrt[]{3}}{\sqrt[]{3}} \\ \sin B=\frac{\sqrt[]{3}}{3} \end{gathered}[/tex]By definition:
[tex]\cos (angle)=\frac{\text{adjacent}}{\text{hypotenuse}}[/tex]Considering angle A, the adjacent side is 5 and the hypotenuse is 5√3. Substituting with this information, we get:
[tex]\cos A=\frac{5}{5\sqrt[]{3}}=\frac{\sqrt[]{3}}{3}[/tex]Considering angle B, the adjacent side is 5√2 and the hypotenuse is 5√3. Substituting with this information, we get:
[tex]\cos B=\frac{5\sqrt[]{2}}{5\sqrt[]{3}}=\frac{\sqrt[]{6}}{3}[/tex]