Answer:
[tex]\phi=\theta a2[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
In the figure ABCD
[tex]m<ADC=\theta a1[/tex]
[tex]m<DAB=90\°[/tex]
[tex]m<ABC=90\°[/tex]
[tex]m<DCB=90\°+\phi[/tex]
Remember that the sum of the internal angles in a quadrilateral is equal to [tex]360\°[/tex]
so
[tex]\theta a1+90\°+90\°+90\°+\phi=360\°[/tex]
[tex]\theta a1+\phi=90\°[/tex] -----> complementary angles
and
[tex]\theta a1+\theta a2=90\°[/tex] -------> complementary angles too
so
[tex]\phi=\theta a2[/tex]
