First we are going to calculate the side BC using the Pythagorean theorem:
[tex]\begin{gathered} BC=\sqrt[]{10^2+20^2} \\ BC=\sqrt[]{300} \\ BC=10\cdot\sqrt[]{3} \end{gathered}[/tex]From the graph we can determine that:
[tex]\measuredangle C=90º[/tex]For the calculation of the angle B we will use the sine of the angle:
[tex]\begin{gathered} sinB=\frac{opposite}{hippotenuse} \\ sinB=\frac{10}{20} \\ \measuredangle B=\sin ^{-1}(0.5) \\ \measuredangle B=30º \end{gathered}[/tex]To find the angle A, we use the law of internal angles of a triangle, which says that the sum of the internal angles of a triangle gives 180º
[tex]\begin{gathered} \measuredangle A+\measuredangle B+\measuredangle C=180º \\ \measuredangle A=180º-30º-90º \\ \measuredangle A=60º \end{gathered}[/tex]: