PQRS is a cyclic quadrilateral. Given ∠QPS = 73°, ∠PQS = 55° and

1.	PQRS is a cyclic quadrilateral. Given ∠QPS = 73°, ∠PQS = 55° and ∠PSR = 82°, calculate:(i) ∠QRS (ii) ∠RQS (iii) ∠PRQ
Answer» (i) Since PQRS is a cyclic quadrilateral ∠QPS + ∠QRS – 180° ⇒ 73° + ∠QRS = 180° ⇒ ∠QRS = 180° – 73° ∠QRS = 107° (ii) Again, ∠PQR + ∠PSR = 180° ∠PQS + ∠RQS + ∠PSR = 180° 55° – ∠RQS + 82° = 180° ∠RQS = 180° – 82° – 55° = 43° (iii) In ∆PQS, by using angles sum property of a ∆. ∠PSQ + ∠SQP + ∠QPS = 180° ∠PSQ + 55° + 73° = 180° ∠PSQ = 180° – 55° – 73° ∠PSQ = 52° Now, ∠PRQ = ∠PSQ = 52° [Oop. ∠s of the same segment] Hence, ∠QRS = 107°, ∠RQS = 43° and ∠PRQ = 52°

PQRS is a cyclic quadrilateral. Given ∠QPS = 73°, ∠PQS = 55° and ∠PSR = 82°, calculate:(i) ∠QRS (ii) ∠RQS (iii) ∠PRQ

Answer»

(i) Since PQRS is a cyclic quadrilateral

∠QPS + ∠QRS – 180°

⇒ 73° + ∠QRS = 180°

⇒ ∠QRS = 180° – 73°

∠QRS = 107°

(ii) Again, ∠PQR + ∠PSR = 180°

∠PQS + ∠RQS + ∠PSR = 180°

55° – ∠RQS + 82° = 180°

∠RQS = 180° – 82° – 55° = 43°

(iii) In ∆PQS, by using angles sum property of a ∆.

∠PSQ + ∠SQP + ∠QPS = 180°

∠PSQ + 55° + 73° = 180°

∠PSQ = 180° – 55° – 73°

∠PSQ = 52°

Now, ∠PRQ = ∠PSQ = 52° [Oop. ∠s of the same segment]

Hence, ∠QRS = 107°, ∠RQS = 43° and ∠PRQ = 52°