Given,
△PQR and
△XYZ:
Side RP = Side ZX = 6 cm
∠QRP =
∠YZX = 140°
Side QR = Side YZ = 9 cm
∴ By SAS criterion for similarity of triangles,
△PQR
∼△XYZ
So,
∠RQP =
∠ZXY = 25° (corresponding parts)
In triangle,
△PQR,
∠PQR = 180° - (
∠RQP +
∠QRP)
⇒ ∠PQR = 180° - (25° + 140°)
⇒ ∠PQR = 15°