Having
used the coordinates of P, Q, S, T as variables, solved
simultaneous quadratics and struggled with decimals for some
time, I got the answer 399.999 and guessed that something
was afoot.
Here’s a shorter way (much fun – thanks KS).

Denote the lengths |RQ| and |RP| by a and b respectively.

By Pythagoras,a^{2} + b^{2} = 60^{2}(1)

Let G be the intersection of the medians, so that GQ and GS
have gradients 2 and 1 respectively. Using these as tangent
ratios: tan/QGS = (2 – 1)/(1 +
2*1) = 1/3(2)

From triangle QGS:/GSR = /QGS
+ /SQG(exterior angle…)