Question

The difference between an exterior angle of
(n-1) sided regular polygon and an exterior
angle of (n + 2) sided regular polygon is 6º.
Find the value of n.​

Answer 1

Answer:

$\sf \: ► \large \: GIVEN : -$

• First polygon having sides (n-1)
• Second polygon having side (n+2)

$\sf \: ► \large \: TO \: \: FIND : -$

• Value of. (n)

$\sf \: ► \large \: SOLUTION : -$

• It is given that polygon is regular then..
• Sum of the exterior angles are equal to 360°

$\blue{\sf \: Sum \: of \: the \: all \: exterior \: angles = 360 \degree}$

$\sf \: For \: first \: polygon \: having \: sides \: (n-1)$

$\sf \: THEN \: one \: exterior \: angle \: of \: (n-1) \: side$

$\sf \: \: one \: exterior \: angle = \frac{360}{n - 1}$

$\sf \: Similarly \: for \: the \: side \: (n+2)$

$\sf \: one \: exterior \: angle \: = \frac{360}{n + 2}$

$\sf \: Now \: According \: to \: the \: Question -$

$\sf \: Difference \: of \: exterior \: angles \: is \: 6°$

$\sf \implies \: \frac{360}{(n - 1)} - \frac{360}{(n + 2)} = 6$

$\sf \implies \: 360 \bigg(\frac{1}{(n - 1)} - \frac{1}{(n + 2)} \bigg) = 6$

$\sf \implies \: 360 \bigg(\frac{(n + 2) - (n - 1)}{(n - 1)(n + 2)} \bigg) = 6$

$\sf \implies \: 360 \bigg(\frac{3}{(n - 1)(n + 2)} \bigg) = 6$

$\sf \implies \: (n - 1)(n + 2) = \frac{360 \times 3}{6}$

$\sf \implies \: (n - 1)(n + 2) = 180$

$\sf \implies \: {n}^{2} - n + 2n - 2 = 180$

$\sf \implies \: {n}^{2} + n - 182 = 0$

$\sf \implies \: {n}^{2} + 14n - 13n - 182 = 0$

$\sf \implies \: n( n+ 14) - 13(n + 14) = 0$

$\sf \implies \: ( n+ 14)(n - 13) = 0$

$\sf \: n = 13 \: \: or \: \: n = - 14$

$\sf \: But \: sides \: can't \: be \: negative$

$\sf \: \boxed{ \sf n = 13}$