Answer:
In the above figure ,
∠QOR=2×∠QMR=100
0
......(1) (Angle made by chord on the major segment is half of angle made by it on centre .)
In △QOR,OQ=OR→∠OQR=∠ORQ........(2)
Using (1) , (2) and angle sum property of triangle in △QOR
∠OQR=∠ORQ=40
0
.................(3)
Now ∠MQP is the exterior angle of △MQR
So, ∠MQP=50
0
+40
0
+40
0
=130
0