Question

A circle is completely divided into "n"sectors in such a way that the angles at the centre of circle (sectional angles) are in ap .if the smallest angle is 8 degree and the largest is 72 degree,calculate n and the angle of the 3 rd sector taking the smallesy angle as first

Answer 1

First angle, a = 8°
nth angle, an = 72°
let the common difference = d

sum of angles = 360°
⇒ (n/2)[a + an] = 360
⇒ (n/2)[8+72] = 360
⇒ (n/2)[80] = 360
⇒ 40n = 360
⇒ n = 360/40
⇒ n = 9

72 = 8+(n-1)d
⇒ 72 - 8 = (9-1)d
⇒ 64 = 8d
⇒ d = 64/8
⇒ d = 8°

a₃ = a + (n-1)d
⇒ a₃ = 8 + (3-1)×8
⇒ a₃ = 24

n = 9
angle of third sector = 24° 

Answer 2

Answer:

n = 9

angle of third sector = 24°

Step-by-step explanation:

First angle, a = 8°

nth angle, an = 72°

let the common difference = d

sum of angles = 360°

⇒ (n/2)[a + an] = 360

⇒ (n/2)[8+72] = 360

⇒ (n/2)[80] = 360

⇒ 40n = 360

⇒ n = 360/40

⇒ n = 9

72 = 8+(n-1)d

⇒ 72 - 8 = (9-1)d

⇒ 64 = 8d

⇒ d = 64/8

⇒ d = 8°

a₃ = a + (n-1)d

⇒ a₃ = 8 + (3-1)×8

⇒ a₃ = 24

n = 9

angle of third sector = 24°