Answer:
Option D. [tex]40\°[/tex]
Step-by-step explanation:
we know that
The inscribed angle measures half that of the arc comprising
so
[tex]m<QPS=\frac{1}{2}(minor\ arc\ QS)[/tex]
we have
[tex]m<QPS=20\°[/tex] ----> given problem
substitute
[tex]20\°=\frac{1}{2}(minor\ arc\ QS)[/tex]
[tex]40\°=(minor\ arc\ QS)[/tex]
Rewrite
[tex]minor\ arc\ QS=40\°[/tex]
