Question

Find the sum of all the angles at the five vertices of the adjoining star.

Answer 1

The adjoining star contains a regular pentagon.

So, sum of interior angles $=(5-2)\times {180}^o={540}^o$

In regular polygon, each angle is of same measure

$\therefore$ Each interior angle measures ${108}^o$

It's supplementary angle measures ${180}^o-{108}^o={72}^o$

Now the triangles formed by the star are isosceles, so the othere angle at the base measures ${72}^o$

$\therefore$ measure of an angle at the vertex will be ${180}^o-{72}^o-{72}^o={36}^o$

$\therefore$ Each angle at the vertices of star has a measure of ${36}^o$.

Thus, the sum of all the angle at five vertices of the star $=5\times {36}^o={180}^o$

Hence, the answer is ${180}^o$.