This is the best for me. killer might have better idea.

ok i solved it to 16/9 but it took me 10 minutes to do this

Here's how :

area of PQS/ area of RQS = (1/2) PS*QS/ (1/2)SR*QS = PS/SR ---eqn 1

PQ + QR + PR = 60

PQ + QR = 60 - PR ------eqn 2

1/2 * PQ *QR = 1/2 * PR * QS

solving and substituting the value of QS we get

PQ * QR = 12 . PR ....eqn 3

now, (PQ + QR )^2 = PQ^2 + QR^2 + 2. PQ.PR -----eqn 4

in eqn 4 substitute the value of PQ + QR from eqn 2 and PQ*QR from eqn 3 and you will get

(60 - pr)^2 = PR^2 + 24 PR

solve the above equation to find

PR = 25

substitute the value of PR in eqn 2 and eqn 3

eqn 2 becomes :

PQ + QR = 35

eqn 3 becomes :

PQ * QR = 300

solving the above two equations :

PQ.(35 - PQ) = 300

PQ^2 -35PQ +300 = 0

PQ = 20,15

so, QR = 20 or 15

but given that side PQ > QR

so, PQ = 20

QR = 15

NOW YOU HAVE ALL THE 3 SIDES OF TRIANGLE as :

PR = 25

PQ = 20

QR = 15

PS = SQRT (PQ^2 - QS^2) = 16

SR = SQRT (QR^2 - QS^2) = 9

PS/SR = 16/9