- Thread starter Faye
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Please show us any work you've been able to do with this problem.

As a suggestion: You don't really need the whole diagram. Consider a regular pentagon. How do you find the interior angles? How do you find the interior angles of a hexagon? Then put the numbers in the diagram and see what you've got.

If you have problems with this, show us your work and we can help you find any flaws.

-Dan

the inside angles of a regular polygon?

If so, go here:

www.mathsisfun.com

And yes, I missed a few of my geometry classes and I am trying to catch up. This was one of the assignments from the class I missed. After looking at this I realized what you are saying. I know the interior angles of a triangle add up to 180. And, the pentagon on the picture add up to 540(each side is 108), and the angles of the hexagon add up to 720(each angle is 120 because they are equal). Also, I know all the sides are congruent that are labeled and I know both triangles are 90 degree triangles. But I still don’t understand how to find the angle of BAC. I believe the answer is 30 degrees because It looks like a 30-60-90 triangle to me., but I have to explain how I came up with thirty degrees and I’m not positive that is correct?

the inside angles of a regular polygon?

If so, go here:

## Interior Angles of Polygons

www.mathsisfun.com

- Joined
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\(\displaystyle \theta_n+\frac{360^{\circ}}{n}=180^{\circ}\)

Hence:

\(\displaystyle \theta_n=180^{\circ}-\frac{360^{\circ}}{n}=\frac{180^{\circ}(n-2)}{n}\)

Let \(\beta=\measuredangle BAC\)...then we know:

\(\displaystyle 2\beta+\theta_5+\theta_6=360^{\circ}\)

\(\displaystyle 2\beta+\frac{180^{\circ}(5-2)}{5}+\frac{180^{\circ}(6-2)}{6}=360^{\circ}\)

Can you finish?

Surrounding point A are 4 angles: see that?

The 2 angles above the triangle are 108 and 120; ok?

The 4 angles total 360; ok?

So the other 2 angles total 360 - 120 - 108 = 132; ok?

Since these 2 angles are equal, then they each equal 66; got that?

And since angle BAC is one of them, then....?

Surrounding point A are 4 angles: see that?

The 2 angles above the triangle are 108 and 120; ok?

The 4 angles total 360; ok?

So the other 2 angles total 360 - 120 - 108 = 132; ok?

Since these 2 angles are equal, then they each equal 66; got that?

And since angle BAC is one of them, then....?

Yes, thank you so much!

\(\displaystyle \theta_n+\frac{360^{\circ}}{n}=180^{\circ}\)

Hence:

\(\displaystyle \theta_n=180^{\circ}-\frac{360^{\circ}}{n}=\frac{180^{\circ}(n-2)}{n}\)

Let \(\beta=\measuredangle BAC\)...then we know:

\(\displaystyle 2\beta+\theta_5+\theta_6=360^{\circ}\)

\(\displaystyle 2\beta+\frac{180^{\circ}(5-2)}{5}+\frac{180^{\circ}(6-2)}{6}=360^{\circ}\)

Can you finish?

Angle BAC equals 66. Thanks a lot!

Surrounding point A are 4 angles: see that?

The 2 angles above the triangle are 108 and 120; ok?

The 4 angles total 360; ok?

So the other 2 angles total 360 - 120 - 108 = 132; ok?

Since these 2 angles are equal, then they each equal 66; got that?

And since angle BAC is one of them, then....?