Oq is a diagonal of quadrilateral nopq. if no=pq, nqo=7y-37, and poq=2y+93, find poq such that nopq is a parallelogram. a.26° b.63° c.97° d.145°

This is the concept of geometry, given that NOPQ is a parallelogram, then it implies that NQO=POQ
hence:
7y-37=2y+93
collecting the like terms we get:
7y-2y=93+37
5y=130
thus;
y=130/5
y=26
thus the value of POQ will be:
POQ=2y+93
=2*26+93
=145
the answer is D. 145°