Question 225075
How can you find the sum of the interior angles of the points of a 5 point star made out of three triangles, but forming a polygon? There are no equilateral triangles and no right angles.
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I'm not sure what you mean.  Here's a 5-pointed star, but I don't
know how to make a 5-pointed star out of three triangles.

Did you mean this kind of 5-pointed star, also known as a pentagram?  

{{{drawing(350,350,-3,3,-3,3,

line(0,2.618033989,.5877852523,.8090169944),
line(.5877852523,.8090169944,2.4898983,.80901699),
line(2.4898983,.80901699,.9510565163,-.3090169944),
line(.9510565163,-.3090169944,1.5388468,-2.118034),
line(1.5388468,-2.118034,0,-1),

line(0,2.618033989,-.5877852523,.8090169944),
line(-.5877852523,.8090169944,-2.4898983,.80901699),
line(-2.4898983,.80901699,-.9510565163,-.3090169944),
line(-.9510565163,-.3090169944,-1.5388468,-2.118034),
line(-1.5388468,-2.118034,0,-1) )}}}

If so, we'll extend the lines like this:

{{{drawing(350,350,-3,3,-3,3,
line(0,2.618033989,1.5388468,-2.118034),
line(0,2.618033989,-1.5388468,-2.118034),
line(2.4898983,.80901699,-2.4898983,.80901699),
line(-2.4898983,.80901699,1.5388468,-2.118034),
line(2.4898983,.80901699,-1.5388468,-2.118034) )}}}

In the center is a 5-sided regular polygon (a regular
pentagon).  The sum of the interior angles of a polygon
is gotten by the formula:

SUM OF INTERIOR ANGLES = (NUMBER OF SIDES - 2) * 180°

For a 5-sided polygon (pentagon) this is (5-2)*180° = 3*180°=540°

Since all 5 angles of a regular pentagon are equal, each
interior angle of the regular pentagon is {{{540/5}}}° = 108°

So I'll mark one of the 108° interior angles of the pentagon:

{{{drawing(350,350,-3,3,-3,3,
line(0,2.618033989,1.5388468,-2.118034),
line(0,2.618033989,-1.5388468,-2.118034),
line(2.4898983,.80901699,-2.4898983,.80901699),
line(-2.4898983,.80901699,1.5388468,-2.118034),
line(2.4898983,.80901699,-1.5388468,-2.118034),
locate(-.6,.7,"108°")
 )}}}

Its suppplement is found by subtracting 180°-108°=72°.
We'll mark it 72°:

{{{drawing(350,350,-3,3,-3,3,
line(0,2.618033989,1.5388468,-2.118034),
line(0,2.618033989,-1.5388468,-2.118034),
line(2.4898983,.80901699,-2.4898983,.80901699),
line(-2.4898983,.80901699,1.5388468,-2.118034),
line(2.4898983,.80901699,-1.5388468,-2.118034),
locate(-.6,.7,"108°"), locate(-.45,1.1,"72°")
 )}}}

That 72° angle is one of the base angles of an isosceles
triangle.  So we'll mark the other base angle 72° also.

{{{drawing(350,350,-3,3,-3,3,
line(0,2.618033989,1.5388468,-2.118034),
line(0,2.618033989,-1.5388468,-2.118034),
line(2.4898983,.80901699,-2.4898983,.80901699),
line(-2.4898983,.80901699,1.5388468,-2.118034),
line(2.4898983,.80901699,-1.5388468,-2.118034),
locate(-.6,.7,"108°"),locate(-.45,1.1,"72°")
locate(0,1.1,"72°") )}}}

Now we can find the angle at the top point of the star by
adding the two equal base angles and subtracting from 180°.

72° + 72° = 144°
180° - 144° = 36°

So each point of the star is 36°.

{{{drawing(350,350,-3,3,-3,3,
line(0,2.618033989,1.5388468,-2.118034),
line(0,2.618033989,-1.5388468,-2.118034),
line(2.4898983,.80901699,-2.4898983,.80901699),
line(-2.4898983,.80901699,1.5388468,-2.118034),
line(2.4898983,.80901699,-1.5388468,-2.118034),
locate(-.6,.7,"108°"),locate(-.45,1.1,"72°")
locate(0,1.1,"72°"), locate(-.16,2,"36°")  )}}}


You wanted the sum of the points interior angles of 
the points.  There are 5 of them, so 5 times 36° is

180°.

Edwin</pre>