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I'm getting arc PTQ is 218 degrees.

Since PQRS is a trapezoid, arcs PS and QR are equal, which leads to the quadratic y^2 + 4y - 60 = 0.  The solutions are y = 6 and y = -10, so we take y = 6.  Then arcs PS and QR are both 78, so arc PTQ is 360 - 78 - 78 = 202 degrees.

Since PQ  is parallel to  RS....then arc PS  = arc QR

So

y^2  + 7y  = 60 - 4y     rearrange as

y^2 - 11y  - 60  =  0      factor

(y - 15) ( y - 4)   =  0

Then y  =15   (reject)   or  y  = 4  (accept)

So  arc PS  = arc QR  =  60 - 4(4)  = 44°

Then arc   PTQ  =  

360  - 2(44)  - 30   =

242°