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 =
![(m<PQT)/(2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/hsrsu2n6goscz9r6lkn5khq85bp520eaej.png)
Plug in the given information and solve for x
4x + 14 =
![(60)/(2)](https://img.qammunity.org/2019/formulas/mathematics/middle-school/6wfq9nwq67laykgsngjby49mn4xi3f1dck.png)
4x + 14 = 30 (subtract 14 on both sides)
4x = 16 (Divide by 4 on both sides)
x = 4
The value of x is 4