Question

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

Answer 1

Hint: Sum of the interior angles of a polygon is${180^ \circ }(n - 2)$. Use this formula.

Given, from the question:
The angles of the pentagon are in the ratio$4:8:6:4:5$,
So, let the respective angles of the pentagon are $4x,8x,6x,4x$and$5x$. Therefore the sum of all angles is:
$ \Rightarrow 4x + 8x + 6x + 4x + 5x = 27x$
And we know that the sum of all the angles of a polygon is ${180^ \circ }(n - 2)$, where $n$is the number of sides in the polygon.
But here we have a pentagon which is a $5$ sided polygon. Thus, in this case the value of $n$ will be $5$.
So using this formula, we’ll get:
\[
   \Rightarrow 27x = {180^ \circ }(n - 2), \\
   \Rightarrow 27x = {180^ \circ }\left( {5 - 3} \right), \\
   \Rightarrow 27x = {540^ \circ }, \\
   \Rightarrow x = 20 \\
\]
The value of $x$ is $20$. So, the measure of respective angles is $4 \times \left( {20} \right),8 \times \left( {20} \right),6 \times \left( {20} \right),4 \times \left( {20} \right)$and$5 \times \left( {20} \right)$.
Thus the angles are ${80^ \circ },{160^ \circ },120,{80^ \circ }$ and${100^ \circ }$.
Note: This question can also be solved as:
The ratio of angles is $4:8:6:4:5$. The sum of ratios is $27$.
And the sum of angles of the polygon is ${180^ \circ }(n - 2)$, in this case it is \[{540^ \circ }\](Putting $n = 5$for pentagon).
So, we have to divide \[{540^ \circ }\]in the ratio$4:8:6:4:5$. Thus the required angles are:
$ \Rightarrow \frac{4}{{27}} \times {540^ \circ },\frac{8}{{27}} \times {540^ \circ },\frac{6}{{27}} \times {540^ \circ },\frac{4}{{27}} \times {540^ \circ }$and$\frac{5}{{27}} \times {540^ \circ }$.
We will get the same result as above.