A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is 8 0 and the largest is 72 0 , calculate n and the angle in the fifth sector.

Question

Philoid · Accepted Answer

A circle is divided into n sectors.

Given,

Angles are in A.P

Smallest angle = a = 8°

Largest angle = l = 72°

Final term of last term of an A.P series is l = a + (n - 1)×d

So,

72° = 8° + (n - 1) ×d

(n - 1) ×d = 64°

Sum of all angles of all divided sectors is

Sum of n terms of A.P whose first term and the last term are known is

Where nis the number of terms in A.P.

So,

n(40°) = 360°

n

n = 9

From equations (1) & (2) we get,

(9 - 1) ×d = 64°

8×d = 64°

d

d = 8°

The circle is divided into nine sectors whose angles are in A.P with a common difference of 8°.

Angle in fifth sector is a + (5 - 1) ×d = 40°

∴n = 9

The angle in the fifth sector = 40°.

A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is 80 and the largest is 720, calculate n and the angle in the fifth sector.

A circle is completely divided into n sectors in such a way that the angles of the sectors are in AP. If the smallest of these angles is 8⁰ and the largest is 72⁰, calculate n and the angle in the fifth sector.