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

Question:

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

(i) pentagon

(ii) octagon

(iii) heptagon

(iv) duodecagon.

Solution:

(i) Sum of the interior angles of the polygon $=(n-2) \pi$

Number of sides in the pentagon $=5$

$\therefore$ Sum of the interior angles of the pentagon $=(5-2) \pi=3 \pi$

Each angle of the pentagon $=\frac{\text { Sum of the interior angles of the polygon }}{\text { Number of sides }}=\frac{3 \pi}{5} \mathrm{rad}$

Each angle of the pentagon $=\left(\frac{3 \pi}{5} \times \frac{180}{\pi}\right)^{\circ}=108^{\circ}$

(ii) Sum of the interior angles of the polygon $=(n-2) \pi$

Number of sides in the octagon $=8$

$\therefore$ Sum of the interior angles of the octagon $=(8-2) \pi=6 \pi$

Each angle of the octagon $=\frac{\text { Sum of the interior angles of the polygon }}{\text { Number of sides }}=\frac{6 \pi}{8}=\frac{3 \pi}{4} \mathrm{rad}$

Each angle of octagon $=\left(\frac{3 \pi}{4} \times \frac{180}{\pi}\right)^{\circ}=135^{\circ}$

(iii) Sum of the interior angles of the polygon $=(n-2) \pi$

Number of sides in the heptagon $=7$

$\therefore$ Sum of the interior angles of the heptagon $=(7-2) \pi=5 \pi$

Each angle of the heptagon $=\frac{\text { Sum of the interior angles of the polygon }}{\text { Number of sides }}=\frac{5 \pi}{7} \mathrm{rad}$

Each angle of the heptagon $=\left(\frac{5 \pi}{7} \times \frac{180}{\pi}\right)^{\circ}=\left(\frac{900}{7}\right)^{\circ}=128^{\circ} 34^{\prime} 17^{\prime}$

(iv) Sum of the interior angles of the polygon $=(n-2) \pi$

Number of sides in the duodecagon $=12$

$\therefore$ Sum of the interior angles of the duodecagon $=(12-2) \pi=10 \pi$

Each angle of the duodecagon $=\frac{\text { Sum of the interior angles of the polygon }}{\text { Number of sides }}=\frac{10 \pi}{12}=\frac{5 \pi}{6}$ rad

Each angle of duodecagon $=\left(\frac{5 \pi}{6} \times \frac{180}{\pi}\right)^{\circ}=150^{\circ}$