Answer:
[tex] \boxed{\stackrel{\Large\frown}{PQS}\:= 222°} [/tex]
Explanation:
Since [tex] \overline{PR} [/tex] is a diameter, arc [tex] \stackrel{\Large\frown}{PQR}\:= 180°. [/tex]
Given that [tex] \angle{SUR} = 42°[/tex] is a subtended central angle by two diameters,
[tex] \stackrel{\large\frown}{RS} \:= 42° [/tex].
Using the rule of arcs, [tex] \stackrel{\Large\frown}{PQR} + \stackrel{\large\frown}{RS} \: = \: \stackrel{\Large\frown}{PQS} [/tex] →
[tex] 180° + \: 42° = \: \stackrel{\Large\frown}{PQS} [/tex]
[tex] 222° = \: \stackrel{\Large\frown}{PQS} [/tex]
[tex] \stackrel{\Large\frown}{PQS} \: = \: 222° [/tex]
