Question

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 8o and the largest is 72o then the angle in the fifth sector is ?

Answer 1

Answer:

Step-by-step explanation:

First term(a) = 8°.

Last term (an) = 72°.

Sum of n terms(Sn) = n/2×(a+an)

(Sum of all angles) 360° =n/2(8°+72°)

360°=n/2(80°)

(2 transferred from RHS to LHS) 360°×2=80°×n

720° = 80°×n

n=720°/80°

n=9.

an=a+(n-1)d

72°=8°+(9-1)d

72°=8°+8d

72°-8°=8d

64°=8d

d=64°/8

d=8°

a5=8°+(5-1)8°

a5=8°+4×8°

a5=8°+32°

a5=40°

Hence, the angle of 5th sector is 40°.

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