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### Sum of the Angles in a Star

```
Date: 09/21/1999 at 11:10:35
From: BonnaIII@aol.com
Subject: Problem

Problem: Find the sum of the measures of the five acute angles that
make up a star.

Sincerely,
Manuel
```

```
Date: 09/21/1999 at 12:55:52
From: Doctor Peterson
Subject: Re: Problem

Hi, Manuel.

I'll assume you have a normal five-pointed star:

+
/ \
/   \
/     \
/       \
+----------------->+------->----------+--------
\         /   start   \       A /   B
\    /             \    /
\               \/
/   \         /   \
/        \  /       \
/        /    \       \
/     /           \     \
/  /                   \  \
+                           +

You can find the sum of the angles easily if you first think not about
the internal angles A, but the external angles B. Imagine you're
taking a walk around the star, following the edges, starting in the
middle of one side, until you get back to the same place. At each
point you will be turning B degrees. How many times will you turn
completely around in your travels? This total turn will be the sum of
all the "B" angles; from that you can find the sum of the "A"s.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/22/1999 at 13:05:49
From: BonnaIII@aol.com
Subject: Re: Problem

Can you please show me how you came up with the answer? I am having a
hard time figuring it out.

I thank you a lot for helping me out.

Manuel
```

```
Date: 09/22/1999 at 13:24:34
From: Doctor Peterson
Subject: Re: Problem

Hi, Manuel.

I'll put a post in the ground in the middle of my star, and attach a
rope to the top. Let's see how many times the rope will wrap around:

/
/
3 /
+ X3
/ \
/   \
\                     /     \
\  X5             /       \               1
+----------------->+------->----------+--------
5   \         /   start   \      Y  /   X1
\    /             \    /
\       *       \/
/   \         /   \
/        \  /       \
/        /    \       \
/     /           \     \
X2 /  /                   \  \
+                           + 4
/    2                        X4 \
/                                     \
\

I first come to point 1, then to 2, then to 3. At that point, I've
gone around once, since I'm in the same direction from the pole as the
place I started. Since I've walked around the pole once, I've turned
360 degrees.

Now I continue to point 4, then point 5, and back to my starting
point, going around the pole a second time. I've now turned a total of
720 degrees.

At each of the five points, I turned some number of degrees X. In
general, these might all be different, so I'll call them X1, X2, X3,
X4, and X5. Do you see why these angles are the amount I turned? At
point 1, I was first pointing to the right along the line I drew, but
I turned to the right until I pointed to 2, so I turned through X
degrees.

If in five turns I turned a total of 720 degrees, then the sum of the
five X's must be 720 degrees.

But each X is the supplement of the acute angle at that point, or
vertex; that is, it's 180 degrees minus the angle Y at that vertex.
This gives me this equation:

(180-Y1) + (180-Y2) + (180-Y3) + (180-Y4) + (180-Y5) = 720

Now you can solve this equation to find what Y1+Y2+Y3+Y4+Y5 is.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 09/22/1999 at 18:01:26
From: BonnaIII@aol.com
Subject: Re: Problem

Thanks a lot for your help!

Manuel
```
Associated Topics:
Middle School Algebra
Middle School Geometry
Middle School Triangles and Other Polygons

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