# Triangles and Polygon measurements

#### topsquark

##### Full Member
Since you didn't actually ask the question I'll take the reasonable assumption that you are struggling with finding the angles.

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

#### Faye

##### New member
So I come up with 45 degrees, but I’m not sure how to show the steps? We’ve been working on 45-45-90 and 30-60-90 triangles. I know the interior angles have to add up to 180 but I don’t understand how to verify that 45 degrees is the correct answer. I’m new to this forum. Sorry I didn’t post right.

#### Denis

##### Senior Member
Are you saying you have not been taught how to calculate
the inside angles of a regular polygon?
If so, go here:

• Faye

#### Faye

##### New member
Are you saying you have not been taught how to calculate
the inside angles of a regular polygon?
If so, go here:
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?

#### MarkFL

##### Super Moderator
Staff member
Suppose we decompose a regular polygon having $$n$$ sides into $$n$$ isosceles triangles. If we look at just one of these triangles, we know the sum of the two equal angles (which will be equal to an interior angle $$\theta_n$$) and the other angle is $$180^{\circ}$$:

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

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?

• Faye

#### Faye

##### New member
Yes, thank you so much!

#### Denis

##### Senior Member
Right! Angles in pentagon = 108, in hexagon = 120.
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....?

• Faye

#### Faye

##### New member
Right! Angles in pentagon = 108, in hexagon = 120.
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....?
Suppose we decompose a regular polygon having $$n$$ sides into $$n$$ isosceles triangles. If we look at just one of these triangles, we know the sum of the two equal angles (which will be equal to an interior angle $$\theta_n$$) and the other angle is $$180^{\circ}$$:

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

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?
Yes, thank you so much!

#### Faye

##### New member
Right! Angles in pentagon = 108, in hexagon = 120.
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....?
Angle BAC equals 66. Thanks a lot!

• MarkFL