Are You An Angle-Star?

Inside pentagon $FGHIJ$ the sum of all the angle is $540^\circ$ . And exterior angles here will be $72^\circ$ and therefore each isosceles triangle will have $\angle A, \angle B, \angle C, \angle D, \angle E$ of $36^\circ$ each. Hence $\angle A +\angle B +\angle C+\angle D+\angle E=36^\circ \times 5 =180^\circ .\square$

A, B, C, D, E are 5 points on a circle. Angle A slits arc DC, Angel B slits arc ED. The same for other angles. We call O the central point of the circle. Angle A = 1/2 Angle DOC, Angle B = 1/2 Angle EOD. The same for other angles. Angle A + B + C +D +E = 360 degree/2 = 180 degree.

In triangle AGF angle A + angle G + angle F = 180 . And if we see triangle GDB angle D + angle B = ANGLE AGF . And in triangle FEC angle E + angle C = angle AFG. So we can write as angle A + angle B+ angle C + angle D + angle E =180. So the correct answer is 180.

just hold ur pen along the line BE(plz hold it at middle point of pen),now move it in counter clockwise direction to line BD,in this u swept an angle B(or angle EBD).Now again move counterclock wise to sweep angle D,thn angle A,angleC ,and angle E,Now see ur pen is along line EB in inverted position with what u started,It means u completed 180 degree

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

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

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

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

By exterior angle, $\angle A+\angle C=\angle EGH$ in $\Delta ACH$

and $\angle B+\angle D=\angle EGH$ in $\Delta BGD$ .

Therefore, $\angle A+\angle B+\angle C+\angle D+\angle E=\angle E+\angle EGH+\angle EHG=180^\circ$ .

Many of the solutions given previously assume that the pentagon formed by the figure is a regular pentagon. This is not necessary. It may be shown that ANY five pointed star figure has a tip angle sum, if you will, of 180. Some other solutions didn't bother to explain their reasoning to the rest of us, only leaving equations, which is annoying.

SOLUTION BETWEEN HERE

I don't remember the name of the theorem that would state that (letting the letters at the tips of the stars refer to the angles in the triangles) A=G+F-180 and likewise with he other cases at the star tips It would be easier to just prove it to yourself than looking it up. Begin by labeling the interior angles of a general triangle as well as the ones formed by extending one of it's sides. A little algebra and subtracting from 180 degrees due to being on a line or in a triangle and you're on your way.

Continuing, we can use the above unnamed theorem to substitute for each angle in the angle sum, resulting in a quantity that we can simplify to 2(mG+mH+mI+mJ+mF)-5*180

As the shape in the center is a pentagon, we know the sum of the measures of vertical angles to G H I J and F is 540 degrees. From this it follows that the sum of the measures of G H I J and F is also 540

Substituting we get 2*540-900=180.

AND HERE

While not a rigorous structured proof, this sketch of the process ought to be sufficient for you to make one if you really care to prove without loss of generality that any five point star figure has a "tip angle" sum of 180 degrees. I did assume that all "five point star figure's tip points also form a convex pentagon...or at least that all those meeting what I'm calling a five point star figure..er

DEFINITION A five point star figure shall be taken to mean the figure defined from a set of five co-planer points which could form a CONVEX pentagon in this way; Connect each point to the two other points which are not adjacent to the point in question with a line segment.

As FGHI is a Pentagon, angle GFJ = angle FGH=108 degree. Therefore, angle AGF= angle AFG = 180-108 =72 degree In that case angle GAF or angle A = 180 - 72x2 = 36 degree. Similarly angle A = angle B = angle C= angle D = angle E = 36 degree Thus the sum of angle A +angle B + angle C + angle D + angle E = 36 X 5 = 180 degree

now GFJIH is a pentagon and we know Interior angle of pentagon is 108. Now consider for angle A .... Now angle HGF =108,so AGF=180-108=72. Similarly AFG=180-JFG=72. Hence angle A=180- angle G - angle F= 180-72-72 =36. Similarly for angles B,C,D,E....... SO 36+36+36+36+36 =180.

Dentro há um pentágono que possui 108 graus cada angulo interno. você subtrai 180, que da 72 logo 2 angulos do triangulo A G F é 72. 72 + 72 = 144. 180 - 144 = 36. logo o angulo de A = 36 multiplicando por 5 = 180

Solution 1.

The sum of the exterior angles of pentagon FGHIJ is

AFG+BJF+CIJ+DHI+EGH = 360 = AGF+BFJ+CJI+DIH+EHG,

(AFG+AGF)+(BJF+BFJ)+(CIJ+CJI)+(DHI+DIH)+(EGH+EHG) = 720,

(180-A)+(180-B)+(180-C)+(180-D)+(180-E) = 720,

A+B+C+D+E = 180.

Solution 2.

(JFG+E+C)+(FGH+B+D)+(GHI+A+C)+(HIJ+B+E)+(IJF+A+D) = 5*180 = 900,

2(A+B+C+D+E)+(JFG+FGH+GHI+HIJ+IJF) = 900,

2(A+B+C+D+E)+180(5-2) = 900,

A+B+C+D+E = 180.

The middle is a pentagon. The sum of the interior angles of a pentagon is 540, since there are 5 sides- (plug into 180 * (n-2)). That means that one angle in the pentagon is 108 degrees. Since with any of these angle in the pentagon continue on a straight line to one of the tips or the pentagon, we can do 180-108=72, which is equal to angles f, g,h,i, and j. Since the tip is a triangle, and there are two of those angles for each tip, we can do 180-72 2=36, which is the tip angle. And since there are 5 tip angles, a,b,c,d, and e, we can do 36 5 to give us 180. Thus the answer is 180.

36 . 5 = 180

There are 5 triangles each with 180 degree the total sum of its 3 angles.if we choose a triangle out of the five say AGF we can see G and F are exterior angles of pentagon GFJIH thus all exterior angles taken in clockwise direction adds up to 360 and another round adds anticlockwise to another 360.The sum of angles of 5 triangles together makes 900.thus adding A+B+C+D+E is like adding all angles of triangles and subtracting the sum of 10 exterior angles ie 720 from it.thus answer is 900-720=180.

GFJIH is a pentagon. its angles are equal, each=108.

$\angle$ BFJ=72= $\angle$ BJF. $\angle$ B=36. so the sum of $\angle$ A, $\angle$ B, $\angle$ C, $\angle$ D, $\angle$ E=180 .

we know the property that the exterior angles of any polygon add up to two straight angles=360 degree. we can see that there are exterior set of angles. therefore, total =2*360=720. and we can see that there are 5 triangles. so, total measure=5 X 180=900 degree. now, this the remaining angles we are asked for=900-720=180