Question

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

$\left(iii\right)$ Heptagon

Answer 1

General formula for interior angles of regular polygon with $n$ sides

$=\left(\frac{n-2}{n}\right)\times {180}^{\circ }$

As we know Heptagon has $7$ sides, so $n=7$

Magnitude of interior angle in degrees

$=\left(\frac{7-2}{7}\right)\times {180}^{\circ }$

$=\frac{{900}^{\circ }}{7}$

$=128{\frac{4}{7}}^{\circ }$

We know $1^{\circ }={60}^{\prime }$, then

${\frac{4}{7}}^{\circ }={\frac{4\times 60}{7}}^{\prime }=\frac{{240}^{\prime }}{7}=34{\frac{2}{7}}^{\prime }$

We know $1^{\prime }={60}^{\prime \prime }$, then

${\frac{2}{7}}^{\prime }=\frac{{120}^{\prime \prime }}{7}\approx {17}^{\prime \prime }$

So,

$128{\frac{4}{7}}^{\circ }={128}^{\circ }{34}^{\prime }{17}^{\prime \prime }$

And, magnitude of interior angle in radians

$=\left(\frac{7-2}{7}\right)\times {\pi }^c$

$[\because {180}^{\circ }={\pi }^c]$

$={\left(\frac{5\pi }{7}\right)}^c$

Hence, magnitude, in degrees and radians are

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