Question

The interior angles of a pentagon are in the ratio $4:8:6:4:5$ Find each angle of the pentagon.

Answer 1

Let the interior angles of the pentagon be $4x,8x,6x,4x,5x$.

Their sum $=4x+8x+6x+4x+5x=27x$

Sum of interior angles of a polygon$=(n-2){180}^{\circ }=(5-2){180}^{\circ }={540}^{\circ }$

$\therefore 27x={540}^{\circ }$

$\Rightarrow \frac{{540}^{\circ }}{27}={20}^{\circ }$

$\therefore$ Angles are

$4x=4\times {20}^{\circ }={80}^{\circ }$

$8x=8\times {20}^{\circ }={160}^{\circ }$

$6x=6\times {20}^{\circ }={120}^{\circ }$

$4x=4\times {20}^{\circ }={80}^{\circ }$

$5x=5\times {20}^{\circ }={100}^{\circ }$