Question

please solved this question

Answer 1

1
the ang tsr is 60° bec it is eq triangle which is equal to ang trs
ang rsp=ang srq=90°
bec they are corner of square
so add that
90+60=ang tsp=ang trq
hence prooved

Answer 2

lets take first, equiletral ∆SRT
1 ) angle tsr =srt=rts = 60°
2) ts =sr= rt

now square spqr
3) all sides equall
sp =rq
4) all angles = 90°
angle psr = angle srq


now ,
ts = tr (from 1 )

sp = rq ( from 3 )

angle tsr+angle rsp = angle trs+ angle srq
60°+90° = 60° +90°


5) hence , ∆TSP = ∆ TRQ

now ,
pt = tq ( from point 5 )


now, taking obtus ∆TSP

sum of angles of triangle = 180°
angle tsp + angle spt + angle pts = 180°
150+ 2 angle spt = 180
2 angle spt = 30
angle spt = 15°