Snitiser7idn
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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.

PQ And QR Are 2 Sides Of A Regular 10sided Polygon Can Someone Please Help Ive Been Struggling On This For Quite A While class=

Sagot :

Xenia168idn

Answer:

Step-by-step explanation:

Measure of inner angle of the n-sided polygon is

[tex]\frac{180(n-2)}{n}[/tex]

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

m∠PQR = [tex]\frac{180(10-2)}{10}[/tex] = 144°

PQ = QR ⇒ m∠QPR = m∠QRP

m∠PRQ = ( 180° - 144° ) ÷ 2 = 18 °

Semsee45idn

Answer:

∠PRQ = 18°

Step-by-step 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°

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