t=46°
u=134°
v=23°
w=23°
Answer:
Solution given;
v=w [base angle of triangle]
<SPQ+<SRQ=180°[sum of opposite angle of a cyclic quadrilateral is 180°]
81+v+53+w=180°
134°+v+v=180°
2v=180°-134°
v=46/2
v=23°
w=v=23°
In triangle ∆ PQR
v+w+u=180°[sum of interior angle of a triangle]
23+23+u=180°
u=180°-46°
u=134°
again
In triangle ∆ PSR
t+81+53=180°
t=180°-134°
t=46°