THEOREM
SUM OF ANGLES IN A TRIANGLE = 180 degrees
[tex]\begin{gathered} A+B+C=180^0 \\ 58+x+x=180^0 \\ 58+2x=180^0 \\ subtract\text{ 58 from bothsides } \\ 58-58+2x=180-58 \\ 2x=122 \\ \\ divide\text{ bothsides by 2} \\ \\ \frac{2x}{2}=\frac{122}{2} \\ \\ x=61^0 \end{gathered}[/tex]
Final answer is 61 degrees
