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Mathematics

If the difference between an interior angle of a regular polygon of (n + 1) sides and an interior angle of a regular polygon of n sides is 4°; find the value of n.

Also, state the difference between their exterior angles.

Rectilinear Figures

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Answer

An interior angle of (n + 1) sided regular polygon = 180°((n+1)−2)(n+1)\dfrac{180°((n + 1) - 2)}{(n + 1)}

An interior angle of n sided regular polygon = 180°(n−2)n\dfrac{180°(n - 2)}{n}

⇒180°((n+1)−2)(n+1)−180°(n−2)n=4°⇒45[((n+1)−2)(n+1)−(n−2)n]=1⇒45[(n−1)(n+1)−(n−2)n]=1⇒45[n(n−1)n(n+1)−(n−2)(n+1)n(n+1)]=1⇒45[n2−nn(n+1)−(n2−2n+n−2)n(n+1)]=1⇒45[n2−nn(n+1)−(n2−n−2)n(n+1)]=1⇒45[n2−n−n2+n+2n(n+1)]=1⇒45[2n(n+1)]=1⇒45×2=n(n+1)⇒90=n2+n⇒n2+n−90=0⇒n2+10n−9n−90=0⇒(n2+10n)−(9n+90)=0⇒n(n+10)−9(n+10)=0⇒(n+10)(n−9)=0⇒n=−10 or 9⇒\dfrac{180°((n + 1) - 2)}{(n + 1)} - \dfrac{180°(n - 2)}{n} = 4°\\[1em] ⇒ 45\Big[\dfrac{((n + 1) - 2)}{(n + 1)} - \dfrac{(n - 2)}{n}\Big] = 1\\[1em] ⇒ 45\Big[\dfrac{(n - 1)}{(n + 1)} - \dfrac{(n - 2)}{n}\Big] = 1\\[1em] ⇒ 45\Big[\dfrac{n(n - 1)}{n(n + 1)} - \dfrac{(n - 2)(n + 1)}{n(n + 1)}\Big] = 1\\[1em] ⇒ 45\Big[\dfrac{n^2 - n}{n(n + 1)} - \dfrac{(n^2 - 2n + n - 2)}{n(n + 1)}\Big] = 1\\[1em] ⇒ 45\Big[\dfrac{n^2 - n}{n(n + 1)} - \dfrac{(n^2 - n - 2)}{n(n + 1)}\Big] = 1\\[1em] ⇒ 45\Big[\dfrac{n^2 - n - n^2 + n + 2}{n(n + 1)}\Big] = 1\\[1em] ⇒ 45\Big[\dfrac{2}{n(n + 1)}\Big] = 1\\[1em] ⇒ 45 \times 2 = n(n + 1)\\[1em] ⇒ 90 = n^2 + n\\[1em] ⇒ n^2 + n - 90 = 0\\[1em] ⇒ n^2 + 10n - 9n - 90 = 0\\[1em] ⇒ (n^2 + 10n) - (9n + 90) = 0\\[1em] ⇒ n(n + 10) - 9(n + 10) = 0\\[1em] ⇒ (n + 10)(n - 9) = 0\\[1em] ⇒ n = -10 \text{ or } 9

Since, number of sides cannot be negative, n = 9.

Exterior angle, when n = 9 = 360°9\dfrac{360°}{9} = 40°

Exterior angle, when (n + 1) = 10 = 360°10\dfrac{360°}{10} = 36°

Difference between their exterior angles = 40° - 36° = 4°

Hence, number of sides, n = 9 and difference between their exterior angles = 4°.

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