Is there a polygon which has only two types of interior angles $120^{o}$ and $60^{o}$ ? If so how many sides such a polygon has?

Open in App

Solution

Here the two interior angles given are $120 °$ and $60 °$
Consider the interior angle $60 °$
The formula for the interior angle of regular pentagon is $\frac{(n - 2) 180^{o}}{n}$
$∴ 60 = \frac{180 n - 360}{n}$
$\Rightarrow 60 n - 180 n = - 360$ $∴ n = 3$
$(n - 2) 180 = 60 k + 120 (n - k)$
$180 = 60 k + 360 - 120 k$
$∴ k = 3$
$∴$ The no. of sides $= 3$
Hence, the answer is $3.$

Suggest Corrections

Similar questions

3240^{o}

The exterior angle of a regular polygon is one – third of its interior angle. How many sides has the polygon?

120^{\circ}

If each interior angle of a regular polygon is $135^{\circ}$ , the polygon has ___ sides.

If the sum of interior angles of a polygon is 1080°, then the polygon has ____ sides.

Join BYJU'S Learning Program