Question

The interior angles of a pentagon are in the ratio​$4:5:6:7:5$. Find each angle​ of the pentagon.​

Answer 1

Given, polygon having interior angles in the ratio​

$4:5:6:7:5$​

Here number of sides of a polygon is $5$. ​

Let the angles be $4n,5n,6n,7n$ and $5n$.​

$\because$ Sum of interior angles of a polygon having n sides $=(n–2)\times {180}^o$​

$\Rightarrow 4n+5n+6n+7n+5n=(5–2)\times {180}^o$ [Since a polygon has $5$ sides]​

$\Rightarrow 27n=3\times {180}^o$ ​

$\Rightarrow 27n={540}^o$ ​

$\Rightarrow n=\frac{540}{27}$

$\Rightarrow n={20}^o$

Then the required angles:
$4\times {20}^o={80}^o$
$5\times {20}^o={100}^o$
$6\times {20}^o={120}^o$
$7\times {20}^o={140}^o$
$5\times {20}^o={100}^o$

Hence, the angles are ${80}^o,{100}^o,{120}^o,{140}^o$ and ${100}^o$ respectively.