Diagram is attached below
m<PQT=60 and m<PQS=4x+14
m<PQS and m<SQT are same
m<PQT = m<PQS + m<SQT = m<PQS + m<PQS = 2m<PQS
m<PQT = 2m<PQS
So m<PQS = [tex]\frac{m<PQT}{2}[/tex]
Plug in the given information and solve for x
4x + 14 = [tex]\frac{60}{2}[/tex]
4x + 14 = 30 (subtract 14 on both sides)
4x = 16 (Divide by 4 on both sides)
x = 4
The value of x is 4
