Question

Find the magnitude, in radians and degrees, of the interior angle of a regular.

(i) pentagon

(ii) octagon

(iii) heptagon

(iv) duodecagon

Answer 1

General formula for interior angles of polygon with n side $=\left(\frac{2n-4}{n}\right)\times {90}^{\circ }$

(i) Pentagon has 5 sides

$\therefore$ magnitude of the interior angle

$=\frac{2\times 5-4}{5}\times {90}^{\circ }$

$=\frac{6}{5}\times 90={180}^{\circ }$

Now,

$\because 1^c=\frac{180}{\pi }$

And each angle of Pentagon

$=\frac{2\times 5-4}{5}\times \frac{\pi }{2}$

$={\left(\frac{3\pi }{5}\right)}^c$ $\therefore {108}^{\circ },{\left(\frac{3\pi }{5}\right)}^{\circ }$

(ii) Octagon

n = 8

$\therefore$ each angle $=\frac{2\times 8-4}{8}\times {90}^{\circ }$

$={135}^{\circ }$

Again,

each angle $=\frac{2\times 8-4}{8}\times \frac{\pi }{2}$

$={\left(\frac{3\pi }{4}\right)}^c$

$\therefore {135}^{\circ }{\left(\frac{3\pi }{4}\right)}^c$

(iii) Heptagon

n = 7

each angle $=\frac{2\times 7-4}{7}\times {90}^{\circ }$

$=\frac{10}{7}\times {90}^{\circ }=\frac{{900}^{\circ }}{7}$

$={128}^{\circ }{34}^{\prime }{17}^{\prime \prime }$

Again,

each angle $=\frac{2\times 7-4}{7}\times \frac{\pi }{2}$

$=\frac{10}{7}\times \frac{\pi }{2}$

$={\left(\frac{5\pi }{7}\right)}^c$ $\therefore {128}^{\circ }{34}^{\prime }{17}^{\prime \prime },{\left(\frac{5\pi }{7}\right)}^c$

(iv) Duodecagon

n = 12

each angle $=\frac{2\times 12-4}{12}\times {90}^{\circ }$

$=\frac{20}{12}\times {90}^{\circ }$

$={150}^{\circ }$

Again,

each angle $=\frac{2\times 12-4}{12}=\frac{\pi }{2}$

$=\frac{20}{12}\times \frac{\pi }{2}$

$={\left(\frac{5\pi }{6}\right)}^c$

$\therefore {150}^{\circ },{\left(\frac{5\pi }{6}\right)}^c$