Question

in the interior angles of an triangle are in A.P in which the first term and common difference are equal. find the measure of bigger angle if the smaller one is 30​

Answer 1

★$\huge\bf{\blue{\underline{Question:-}}}$

The angles of a polygon are in A.P. with common difference 5 degree if the smallest angle is 120 degree find the number of sides of the polygon

Let there be in n sides in the polygon.

————————————————————————————

★$\huge\bf{\red{\underline{Answer:-}}}$

n = 9.

————————————————————————————

★$\huge\bf{\green{\underline{Explanation:-}}}$

sum of all n interior angles of polygon = (n – 2) * 180°

the angles are in A. P. with the smallest angle = 120°

common difference = 5°

∴ Sum of all interior angles of polygon

= n/2[2 * 120 + ( n – 1) * 5

n/2 [2 * 120 + (n – 1) * 5] = (n – 2) * 180

⇒ n/2 [5n + 235] = (n – 2 ) * 180

⇒ 5n2 + 235n = 360n – 720

⇒ 5n2 – 125n + 720 = 0 ⇒ n2 – 25n + 144 = 0

⇒ (n – 16 ) (n – 9) = 0 ⇒ n = 16, 9

Also if n = 16 then 16th angle = 120 + 15 * 5 = 195° > 180°

∴ not possible.

Hence n = 9.

————————————————————————————

Answer 2

Answer:

sum of all n interior angles of polygon = (n – 2) * 180°

the angles are in A. P. with the smallest angle = 120°

common difference = 5°

∴ Sum of all interior angles of polygon

= n/2[2 * 120 + ( n – 1) * 5

n/2 [2 * 120 + (n – 1) * 5] = (n – 2) * 180

⇒ n/2 [5n + 235] = (n – 2 ) * 180

⇒ 5n2 + 235n = 360n – 720

⇒ 5n2 – 125n + 720 = 0 ⇒ n2 – 25n + 144 = 0

⇒ (n – 16 ) (n – 9) = 0 ⇒ n = 16, 9

Also if n = 16 then 16th angle = 120 + 15 * 5 = 195° > 180°

∴ not possible.

Hence n = 9.