Question

5) Angle PQR = 65° is an .......... angle.​

Answer 1

Step-by-step explanation:

(i) ∠QPR 

∴ PQ is a diameter 

∴ ∠PRQ = 900  [Angle in a semi-circle is 900] 

In ΔPQR, 

∠QPR + ∠PRQ + ∠PQR = 1800 [Angle Sum Property of a triangle] 

=> ∠QPR + 900 + 650= 1800

=> ∠QPR + 1550 = 1800

=> ∠QPR = 1800 - 1550 

=> ∠QPR = 250. 

(ii) ∠PRS 

∴ PQRS is a cyclic quadrilateral 

∴ ∠PSR + ∠PQR = 1800 [∴ Opposite angles of a cyclic quadrilateral are supplementary] 

=> ∠PSR + 650 = 1800 

=> ∠PSR = 1800- 650 

=> ∠PSR = 1150 

In DPSR, 

∠PSR + ∠SPR + ∠PRS = 1800 [Angles Sum Property of a triangle] 

=> 1150 + 400 + ∠PRS = 1800 

=> 1150 + ∠PRS = 1800 

=> ∠PRS = 1800 - 1550 

=> ∠PRS = 250 

(iii) ∠QPM 

∴ PQ is a diameter 

∴ ∠PMQ = 900 [∴ Angle in a semi - circle is 900] 

In ΔPMQ, 

∠PMQ + ∠PQM + ∠QPM = 1800 [Angle sum Property of a triangle] 

=> 900 + 500 + ∠QPM = 1800 

=> 1400 + ∠QPM = 1800

=> ∠QPM = 1800 - 1400 

=> ∠QPM = 400.