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**Hint:**We will solve this question by the sum of triangle. We will solve the question with the help of the angle sum property of a triangle which means the sum of all the angles in the triangle is equal to ${180^ \circ }$.

**Complete step-by-step answer:**

Let us first take the $\vartriangle PCE$

Angle 1 is exterior angle so, we use the property of exterior angle here

$\angle 1 = \angle b + \angle d$ …………..1

Now, in $\vartriangle BGD$ angle 2 is exterior angle for $\vartriangle BGD$

$\angle 2 = \angle c + \angle e$ ……………2

In $\vartriangle APG$ we know that sum of all the angle of the $\vartriangle APG$ is equal to ${180^ \circ }$

Angles in $\vartriangle APG$ are $\angle a, \angle 1 and \angle 2$

By angle sum property of a triangle

$\angle a + \angle 1 + \angle 2 = {180^{^ \circ }}$

Now, put the value of angle 1 and angle 2 by the equation 1 and 2

$\angle a + \angle d + \angle c + \angle e + \angle b = {180^ \circ }$ \[\]

So, the sum of the all the angles marked at the corners A,B,C,D,E is ${180^ \circ }$

**So, the correct answer is “Option C”.**

**Note:**In this question always remember that the angle of each exterior point is always the sum of two adjacent interior angles ${180^ \circ }$.

Alternative method: The adjoining star contains a regular pentagon, and we know the sum of interior angles is $ = {540^ \circ }$, and each angle is of the same measures so we can find the particular angle by dividing the sum by 5.

$ = \dfrac{{540}}{5}$

$ = {108^ \circ }$

It`s supplementary angles measure ${180^ \circ } - {108^ \circ } = {72^ \circ }$

The triangle formed by the star are isosceles triangle in which two angle are $ = {72^ \circ }$

So, we have the find the third angle

=${180^ \circ } - {72^ \circ } - {72^ \circ } = {36^ \circ }$

Each angle at the vertices of the star is equal to ${36^ \circ }$. Thus the sum of all the angle of the star

=${5 \times 36^ \circ } = {180^ \circ }$

Hence the answer is ${180^ \circ }$.

Option C is correct.

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