Question 43163
This problem is very if you draw the figure first.
Consider the triangle PTR.
Let p, t and r be the lengths of the sides RT, PR and PT respectively. 
Also, denote the internal abgles of the triangle as P, T and R e.g. < RTP = T.


{{{drawing(400,200,-100,100,-10,90,
line(-90,0,90,0),
line(90,0,-65.28,76.084),
line(-65.28,76.084,-90,0),
locate(-93,0,T),
locate(93,0,P),
locate(-66,85,R),
locate(0,-2,r=180),
locate(30,38,t=unknown),
locate(-75,40,p=80),
locate(-86,15,72^o)
)}}}


Look, you know from the given data that r = 180 m, p = 80 m and T = {{{72^o}}}.
Now you are required to find 't'.
In order to find 't' apply cosine rule.
{{{t^2 = p^2 + r^2 - 2*p*r*cos(T)}}}
or {{{t^2 = 80^2 + 180^2 - 2*80*180*cos(72^o)}}}
or {{{t^2 = 6400 + 32400 - 28800cos(72^o)}}}
or {{{t^2 = 38800 - 28800*0.309}}}
or {{{t^2 = 38800 - 8899.7}}}
or {{{t^2 = 29900.3}}}
or {{{t = sqrt(29900.3)}}}
or {{{t = 172.92}}}


Hence, the required length of PR is 172.92 m.