Question

the 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 4th sector

Answer 1

Answer:

n = 9,  $a_{4}$ = 32°

Step-by-step explanation:

First angle, $a_{1}$ = 8°,  $n_{th}$ angle, $a_{n}$ = 72°, let the common difference be "d"

sum of angles is 360°

⇒ ($\frac{n}{2}$)($a_{1}$ + $a_{n}$) = 360°

⇒ ($\frac{n}{2}$)(8°+72°) = 360°

⇒  ($\frac{n}{2}$)(80°) = 360°

⇒ 40° × n = 360°

⇒ n = 360° ÷ 40°

⇒ n = 9

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

$a_{4}$ = $a_{1}$ + (n - 1)d

⇒ $a_{4}$ = 8° + (4 - 1) × 8°

⇒ $a_{4}$ = 32°

n = 9 sectors and angle of $4^{th}$ sector = 32°