Triangles and Polygon measurements

Faye

New member
Joined
Mar 16, 2019
Messages
6
11397
 

topsquark

Junior Member
Joined
Aug 27, 2012
Messages
245
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
Joined
Mar 16, 2019
Messages
6
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
Joined
Feb 17, 2004
Messages
1,437
Are you saying you have not been taught how to calculate
the inside angles of a regular polygon?
If so, go here:
 

Faye

New member
Joined
Mar 16, 2019
Messages
6
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
Joined
Nov 24, 2012
Messages
994
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

New member
Joined
Mar 16, 2019
Messages
6
Yes, thank you so much!
 

Denis

Senior Member
Joined
Feb 17, 2004
Messages
1,437
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

New member
Joined
Mar 16, 2019
Messages
6
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
Joined
Mar 16, 2019
Messages
6
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!
 
Top