If ABCD is a rectangle, each of its four angles at the vertex is a right angle.
Then the angle at D is a right angle (measure of 90 degrees).
The right angle is divided in two angles, <1 and <2, by the diagonal, so <1 and <2 are complementary.
Then, we can write:
[tex]\begin{gathered} m\angle D=m\angle1+m\angle2=90\degree \\ (5x)+(8x-1)=90 \\ 13x-1=90 \\ 13x=90+1 \\ 13x=91 \\ x=\frac{91}{13} \\ x=7 \end{gathered}[/tex]
Then, with the value of x we can calculate m<1 and m<2:
[tex]\begin{gathered} m\angle1=5x=5\cdot7=35\degree \\ m\angle2=8x-1=8\cdot7-1=56-1=55\degree \end{gathered}[/tex]
Answer:
x = 7
m<1 = 35º
m<2 = 55º
