we know that
An exterior angle is one that has its vertex at an outer point of the circumference.
The measure of the external angle is the semidifference of the arcs that it covers.
in this problem the external angle is 80°
and
the semidifference of the arcs that it covers is ----> (arcSQP-arc PS)/2
so
(arcSQP-arc PS)/2=80°
Let
x-----------> arc SQP
then
arc PS=360-x
(arcSQP-arc PS)/2=80°----->(x-(360-x))/2=80°-----> (x-360+x)=160°
(2x)=160°+360°-------> x=520/2---------> x=260°
therefore
arc arcSQP=260°
arcPS=360-260------> arc PS=100°
see the attached figure to better understand the problem
