From a given right triangle,
We observe that,
[tex]\begin{gathered} O\text{pposite side=6 cm} \\ \text{Adjacent side=6}\sqrt[]{3}cm \\ \text{Hypotenuous side=12cm} \end{gathered}[/tex]
Using the formula,
[tex]\begin{gathered} \sin A=\frac{opp}{hyp} \\ \sin A=\frac{6\sqrt[]{3}}{12} \\ \sin A=\frac{\sqrt[]{3}}{2} \\ A=\sin ^{-1}(\frac{\sqrt[]{3}}{2}) \\ A=60^{\circ} \end{gathered}[/tex]
Using the formula,
[tex]\begin{gathered} \sin B=\frac{opp}{hyp} \\ \sin B=\frac{6}{12} \\ \sin B=\frac{1}{2} \\ B=30^{\circ} \end{gathered}[/tex]
Hence, the answer is,
[tex]A=60^{\circ},B=30^{\circ}[/tex]
