Question

PQ and QR are 2 sides of a regular 10-sided polygon

Can someone please help I've been struggling on this for quite a while.

Answer 1

Explanation:

Measure of inner angle of the n-sided polygon is

`(180(n-2))/(n)`

~~~~~~~~~~~~~~

m∠PQR =
`(180(10-2))/(10)` = 144°

PQ = QR ⇒ m∠QPR = m∠QRP

m∠PRQ = ( 180° - 144° ) ÷ 2 = 18 °

Answer 2

∠PRQ = 18°

Explanation:

Sum of interior angles of a polygon = (n - 2) × 180°
(where n is the number of sides)

Interior angle of a regular polygon = sum of interior angles ÷ n
(where n is the number of sides)

Given:

• n = 10

⇒ Sum of interior angles of a polygon = (10 - 2) × 180° = 1440°

⇒ Interior angle of a regular polygon = 1440° ÷ 10 = 144°

Therefore, as PQ and QR are the sides of the polygon, ∠PQR = 144°

Sum of interior angles of a triangle = 180°

⇒ ∠PQR + ∠RPQ + ∠PRQ = 180°

⇒ 144° + ∠RPQ + ∠PRQ = 180°

⇒ ∠RPQ + ∠PRQ = 36°

As the polygon is regular, PQ = QR which means that ΔPQR is isosceles.

Therefore, ∠RPQ = ∠PRQ

⇒ ∠PRQ = 36° ÷ 2 = 18°