Question 1199664
<pre>
The drawing on the site is not to scale. This one is:

{{{drawing(375,400,-4,11,-1,16,
line(0,8.692867378,3.9615679,0),

red(arc(0,8.692867378,6,-6,90,115),arc(0,8.692867378,6,-6,294.5,319)),

line(-2.8,0,10,0), line(0,8.692867378,0,15),line(0,8.692867378,-2.874325933,15),
line(0,8.692867378,10,0),locate(-3.1,15.68,Q), locate(-.5,8.7,P), locate(4,0,R), locate(10,0,S),locate(8,1,41^o), locate(-.1,15.68,T))}}}

Extend TP to U

{{{drawing(375,400,-4,11,-1,16,
red(arc(0,8.692867378,6,-6,90,115),arc(0,8.692867378,6,-6,294.5,319),
locate(-.1,0,U)
),

line(0,8.692867378,3.9615679,0),red(line(0,0,0,8.692867378)),
line(-2.8,0,10,0), line(0,8.692867378,0,15),line(0,8.692867378,-2.874325933,15),
line(0,8.692867378,10,0),locate(-3.1,15.68,Q), locate(-.5,8.7,P), locate(4,0,R), locate(10,0,S),locate(8,1,41^o), locate(-.1,15.68,T))}}}

∠TPQ = ∠UPR because they are vertical angles.

∠TPQ = ∠SPR = ∠UPR

PR bisects ∠UPS

ΔSUP is a right triangle

∠UPS is complementary to ∠USP = 41<sup>o</sup>

∠UPS = 90<sup>o</sup> - 41<sup>o</sup> = 49<sup>o</sup>

∠RPS is half of ∠UPS, and half of 49<sup>o</sup> is 24.5<sup>o</sup>  

We now know two angles of ΔSRP, 24.5<sup>o</sup> and 41<sup>o</sup>,

So we add them and subtract from 180<sup>o</sup>.

∠SRP = 180<sup>o</sup>-(24.5<sup>o</sup>+41<sup>o</sup>) = 114.5<sup>o</sup>

Edwin</pre>