Question

solve me please and explanation​

Answer 1

Answer:

(i) PQRS is a cyclic quadrilateral. [Given] ∴ ∠PSR + ∠PQR = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

∴ 110° + ∠PQR = 180°

∴ ∠PQR = 180° – 110°

∴ ∠PQR = 70°

(ii)

∠PSR= 1/2 m (arc PQR) [Inscribed angle theorem] 110°= 1/2 m (arcPQR)

∴ m(arc PQR) = 220°

(iii) In ∆PQR, side PQ ≅ side RQ [Given]

∴ ∠PRQ = ∠QPR [Isosceles triangle theorem]

Let ∠PRQ = ∠QPR = x Now, ∠PQR + ∠QPR + ∠PRQ = 180° [Sum of the measures of angles of a triangle is 180°]

∴ ∠PQR + x + x= 180° ∴ 70° + 2x = 180°

∴ 2x = 180° – 70° ∴ 2x = 110°

∴ x = 100°/2 = 55°

∴ ∠PRQ = ∠QPR = 55°….. (i)

But, ∠QPR = 1/2 nm(arc QR) [Inscribed angle theorem] ∴ 55° = 1/2 m(arc QR)

∴ m(arc QR) = 110°

(iv) ∠PRQ = ∠QPR =55° [From (i)]

∴ m ∠PRQ = 55°

HOPE IT HELPS

Thanks