Question

Exterior angle of a regular polygon having n sides is more than
that of the polygon having n sides by 50°. Find the number of
the sides of each polygon.​

Answer 1

Correct Question :

Exterior angle of a regular polygon having n sides is more than that of the polygon having n square sides by 50°. Find the number of the sides of each polygon.

$\rule{200}{2}$

AnswEr :

$\star \:\sf Exterior\:angle\:Having\:n\:sides = \dfrac{360}{n}\\\\ \star \:\sf Exterior\:angle\: Having\:n^2\:sides=\dfrac{360}{n^2}$

$\rule{150}{1}$

$\underline{\bigstar\:\textsf{According to the Question Now :}}$

$:\implies\tt \dfrac{360}{n}-\dfrac{360}{n^2}=50\\\\\\:\implies\tt\dfrac{360n - 360}{n^2}= 50\\\\\\:\implies\tt 360n - 360 = 50n^2\\\\\\:\implies\tt 10(36n - 36) = 10(5n^2)\\\\\\:\implies\tt 36n - 36 = 5{n}^{2}\\\\\\:\implies\tt 5n^{2} - 36n + 36 = 0\\\\\\:\implies\tt5n^{2} - 30n - 6n + 36 = 0\\\\\\:\implies\tt 5n(n - 6) - 6(n - 6) = 0\\\\\\:\implies\tt (n - 6)(5n - 6) = 0\\\\\\:\implies\tt \green{n = 6} \quad or \quad \red{n =\dfrac{6}{5}}$

⠀

$\therefore\:\underline{\textsf{By ignoring fractional value, Sides will be \textbf{6.}}}$

Answer 2

Answer:

ANSWER IS IN ATTACHMENT. ♡