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In the star shape shown in figure, the sum of the angle marked at the corners A,B,C,D,E is?
                    
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A. 90
B. 135
C. 180
D. 140
E. None of these

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Last updated date: 16th Jun 2024
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Answer
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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.