Recall that the diagonals of a rhombus are perpendicular bisectors of each other, therefore all the right triangles formed by the diagonals as shown in the given diagram are congruent, therefore:
[tex]\begin{gathered} \measuredangle1=\measuredangle4=\measuredangle2, \\ \measuredangle3=39^{\circ}, \\ \measuredangle1+39^{\circ}=90^{\circ}. \end{gathered}[/tex]
Solving the last equation for angle 1 we get:
[tex]\begin{gathered} \measuredangle1=90^{\circ}-39^{\circ}, \\ \measuredangle1=51^{\circ}. \end{gathered}[/tex]
Answer:
[tex]\begin{gathered} m\angle1=51^{\circ}, \\ m\angle2=51^{\circ}, \\ m\angle3=39^{\circ}, \\ m\angle4=51^{\circ}. \end{gathered}[/tex]
