Question

In the adjoining fig < POR =110° then <POR adjoining​

Answer 1

Answer:

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Answer 2

Answer:

PQRS is a cyclic quadrilateral. [Given]

∴ ∠PSR + ∠PQR = 180° [Opposite angles of a cyclic quadrilateral are supplementary]

∴ 110° + ∠PQR = 180°

∴ ∠PQR = 180° – 110°

∴ m ∠PQR = 70°

ii. ∠PSR= 1/2 m (arcPQR) [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°