In a rhombus, the diagonals intersect at an angle of 90º.
Let O be the intersection of both diagonals.
Since the diagonals on a parallelogram intersect at the middle point, and all the sides of a rhombus have the same meaure, then all the triangles that are formed (ROS, TOS, TOQ, ROQ) are congruent.
Then, we know that:
[tex]m\angle1=90º[/tex][tex]m\angle2=40º[/tex]
Since the angles 3 and 2 are complementary angles, then:
[tex]m\angle3=50º[/tex]
Since the angles 4 and 3 are congruent, then:
[tex]m\angle4=50º[/tex]
Therefore:
[tex]\begin{gathered} m\angle1=90º \\ m\angle2=40º \\ m\angle3=50º \\ m\angle4=50º \end{gathered}[/tex]
