In the given figure, ΔQPS = ΔSRQ. Find each value.

(a) x

(b) ∠PQS

(c) ∠PSR

(a) ΔQPS = ΔSRQ

∠QPS = ∠SRQ (Congruent part of congruent triangles)

106 = 2x + 12

⇒ 106 – 12 = 2x

⇒ 94 = 2x

⇒ x = 47

∠QRS = 2 × 47 + 12 = 94 + 12 = 106°

So, PQRS is a parallelogram.

∠QSR = 180° – (42° + 106°) = 180° – 148° = 32°

(b) ∠PQS = 32° (alternate interior angles)

(c) ∠PSQ = 180° – (∠QPS + ∠PQS) = 180° – (106° + 32°) = 180° – 138° = 42°

∠PSR = 32° + 42° = 74°