Twinkle Twinkle Little Star

SIMPLE UNREADABLE SOLUTION

Create a point in the middle of the pentagon. Then create 5 identical triangles from this point where the 2 points of the pentagon are their base angles.

360 / 5 = 72 (the angle of the middle point of your triangle).

(180 - 72) : 2 = 54 (the base angle of the triangle: half of the inside angle of the pentagon)

360 - 54 × 4 = 144 (the 2 angles of the star's triangle. Why 4? See previous step.)

And finaly

180 - 144 = 36 (the angle of one of the triangles points, in this case of a, b, c, d, e)

5 × 36 = 180

P.S.

Don't judge me for the unreadability of this solution. It's the first time I write a one. Thank you for reading!

The interior angle formula of the a regular polygon is $(n-2)*180^\circ$ where n is the number of sides, so the sum of the interior angles of the pentagon in the center is $(5-2)*180^\circ=540^\circ$ . It is regular such that the angles are equal so that each interior angle equals $\frac{540^\circ}{5}=108^\circ$ . This lets us find the other two angles of the triangles a, b, c, d, and e since the interior angles of the pentagon and the other angles of the triangles a, b, c, d, and e form straight lines (180 $^\circ$ ) such that the other angles of the triangles a, b, c, d, and e are equal to $180^\circ-108^\circ=72^\circ$ .

Triangles also have 180 degrees from interior angle formula, which determines that the angles a, b, c, d, and e are equal to $180^\circ-2*72^\circ=36^\circ$ . Thus the sum of angles a, b, c, d, and e=(5*36^\circ=180^\circ.

I KNEW IT! Reading Chinese Comics will get me somewhere

Draw $\space \color{#20A900}{\textbf{a circle}}\space$ that touches the 5 heads of the star A,B,C,D,E

Now every single one of them forms an Inscribed angle

$\Large\color{#3D99F6}{Inscribed \space angle} =\frac12 \color{#3D99F6}{it's \space corresponding \space Arc}$

$\color{#D61F06}{\textbf{Thus}}$

$\color{#3D99F6}{\angle A=\frac12 \widehat {CD}}$

$\color{#3D99F6}{\angle B=\frac12 \widehat {DE}}$

$\color{#3D99F6}{\angle C=\frac12 \widehat {EA}}$

$\color{#3D99F6}{\angle D=\frac12 \widehat {AB}}$

$\color{#3D99F6}{\angle E=\frac12 \widehat {BC}}$

And since this is a homogenous star (Assumed just to make it easier to solve quickly)

(hint: contains an $\textbf{Regular pentagon}$ in it's core)

$\color{#D61F06}{\textbf{Then}}$

$\angle A=\angle B=\angle C=\angle D=\angle E$

$\widehat {CD}=\widehat {DE}=\widehat {EA}=\widehat {AB}=\widehat {BC}=\frac{360}{5}=72$

$\angle B=\angle C=\angle D=\angle E=\angle A=\frac12 \times \widehat {72} =36$

$\color{#D61F06}{\textbf{So;}}$

$\angle A+\angle B+\angle C+\angle D+\angle E=5\times \angle A=5 \times 36 =\boxed{\color{#D61F06}{180}}$

It'll always be the same answer

In case if it's not a Regular shaped star

Substitute in the below equation by the first 5 equations

$\widehat {CD}+\widehat {DE}+\widehat {EA}+\widehat {AB}+\widehat {BC}=360$

You will get

$2\angle A+2\angle B+2\angle C+2\angle D+2\angle E=360$

$2(\angle A+\angle B+\angle C+\angle D+\angle E)=360$

$\angle A+\angle B+\angle C+\angle D+\angle E=\frac{360}{2}=\boxed{\color{#D61F06}{180}}$

Is that what you meant? Eli Ross

That was my answer to the same problem

Just for fun, an alternative solution:

Imagine you start at point $a$ and walk along the figure, passing points $c$ , $e$ , $b$ , $d$ , and back to $a$ .

At each point you turn over an angle of $180^\circ-x$ , where $x = a = \cdots = e$ . The total angle through which you turn is $5(180^\circ-x) = 900^\circ-5x$ .

As you do all this, you turn through two four revolutions: a total of $2\cdot 360 = 720^\circ$ . Thus we may write $900^\circ-5x = 720^\circ.$ It follows immediately that $5x = 900 - 720 = 180^\circ$ , and that is the answer we are looking for. (And, of course, $x = 36^\circ$ .)

Note: If we had walked the pentagon instead of the star, we would only turn through one revolution. The equation would then be $900^\circ-5x = 360^\circ$ , from which $x = (900-360)/5 = 108^\circ$ . This is the angle between sides of the pentagon.

Since the figure does not have to be regular in any way, you can modify it for convenience. Bring angles e and b up toward angle a. As you do so, the measures of angles e and b approach zero, and you are left with $\triangle$ ABC with angles adding to 180 degrees.

5 triangles can be made using 2 points of the star and one point of the pentagon. Each point of the pentagon therefore has an angle of 180 - (2 of a/b/c/d/e) degrees. The interior angles of a pentagon add up to 540 degrees, and the 5 points add up to 900 - 2a - 2b - 2c - 2d - 2e.
So 900 - 2(a + b + c + d + e) = 540, so a + b + c + d + e = 180

Th simplest solution is noticing that four sides of the pentagon combined with an arm of the star forms a quadrilateral. Assuming the pentagon is a regular pentagon where each angle is 108 degrees, you can solve for the angle in the arm of the star, denoted as theta.

theta = 360 - 3(108) theta = 36

Multiply theta by 5, and you will get 180.

We know that sum of all exterior angles of the pentagon is 360 and there are 5 triangles on the sides of pentagon. If we think that the base of the triangles are sides of the pentagon ,sum of those all base angles will be [2 sum of exterior angles] which is equal to 360 2 = 720 . Then sum if angles in all 5 triangles is 180*5 = 900 . Then a+b+c+d+e = 900 -720 = 180.

Creat a pentagon by joining the vertices by line segment, divide each vertex angle as 4 equal angles, 1/4 x (540/5) so each is 18, then a=2x18 = 36, b=36, c=36, d==36 & e=36, Then a+b+c+d+e =36 x 5 = 180