Geometry, find an angle - parent needs help

[drchrisdvm2009]

Hi everyone! I'm a parent to some incredibly smart kids. We homeschool, and I'm having trouble finding an answer to a problem. (I'll post a photo). Any help on how to get the solution is greatly appreciated. I think we are CLOSE, but I'm coming up with something different than the answer key.


The question has two isosceles triangles connected by one of their sides (that are equal). We are given a single angle outside of the two triangles, of 60 degrees. We are asked to solve for x, which is the equal angle of a second triangle. How we are going about this is as follows:
1) We know that the left sided triangle has two angles X and the vertex angle is 180-2x.
2) we know that the vertex angle of the second triangle then has to be 120-x
3) If we label the second triangle, the opposite angles can be labeled as Y.
4) This also means that the vertex of the second triangle can also equal 120-2y.
5) The straight line across the top of the two triangles would equal 180-y, or 180-2x.
6) This is where I fail and can't figure out how to bring it home. I THOUGHT I could go y=2x at that point, or X=1/2y and then plug it back in, but I'm ripping my hair out, next to my wife and two children. One kid is trying to do bisecting lines to make right angles, the other is following in the same format I was going.

HELP!

Thanks so much in advance.

Chris

 

[drchrisdvm2009]

I think I got it now. just needed to do more algebra. 180-y = 180-2x; y=2x or x=1/2y. Then plugging it back in, 120-x+y+y=180 --> 120-1/2y+y+y=180 --> 120 + 1.5y = 180 --> 1.5y=60 --> y=40 --> x=1/2(40) --> x=20.

Just had to stare at it for a while. Thanks everyone!

 

[drchrisdvm2009]

My 13yr old came up with a different way, and I wanted to verify that this was appropriate. ( I know there are multiple ways to skin a cat) but he set a line parallel with the base (I'll include a photo), made circles at each point to equal 360 degrees, found that some of the angles must be 120 degrees. Then, because the ratio of the outside angles of the newly formed parallelogram was 2:1, he put the angles of the triangles at a 2:1 ratio, or 3a=60 degrees (he used different letters for variables, not sure why), so a (or x) = 20. Are there situations that this would not work? I don't want him to learn a method that will fail him in the future.



He ALWAYS solves questions differently than I do, and I love it, but it's hard for me to judge if it's appropriate/mathematically sound or not.

 

[Steven G]

Hi everyone! I'm a parent to some incredibly smart kids. We homeschool, and I'm having trouble finding an answer to a problem. (I'll post a photo). Any help on how to get the solution is greatly appreciated. I think we are CLOSE, but I'm coming up with something different than the answer key.

View attachment 33920
The question has two isosceles triangles connected by one of their sides (that are equal). We are given a single angle outside of the two triangles, of 60 degrees. We are asked to solve for x, which is the equal angle of a second triangle. How we are going about this is as follows:
1) We know that the left sided triangle has two angles X and the vertex angle is 180-2x.
2) we know that the vertex angle of the second triangle then has to be 120-x
3) If we label the second triangle, the opposite angles can be labeled as Y.
4) This also means that the vertex of the second triangle can also equal 120-2y.
5) The straight line across the top of the two triangles would equal 180-y, or 180-2x.
6) This is where I fail and can't figure out how to bring it home. I THOUGHT I could go y=2x at that point, or X=1/2y and then plug it back in, but I'm ripping my hair out, next to my wife and two children. One kid is trying to do bisecting lines to make right angles, the other is following in the same format I was going.

HELP!

Thanks so much in advance.

Chris

Please label what you think the angles should be as it is not easy to follow what you are saying. For example, you say in 5) that a straight line has an angle other than 180 (unless you think y=x=0). Also label each vertex.

 

[Harry_the_cat]

My 13yr old came up with a different way, and I wanted to verify that this was appropriate. ( I know there are multiple ways to skin a cat) but he set a line parallel with the base (I'll include a photo), made circles at each point to equal 360 degrees, found that some of the angles must be 120 degrees. Then, because the ratio of the outside angles of the newly formed parallelogram was 2:1, he put the angles of the triangles at a 2:1 ratio, or 3a=60 degrees (he used different letters for variables, not sure why), so a (or x) = 20. Are there situations that this would not work? I don't want him to learn a method that will fail him in the future.

View attachment 33923

He ALWAYS solves questions differently than I do, and I love it, but it's hard for me to judge if it's appropriate/mathematically sound or not.

You say: "Then, because the ratio of the outside angles of the newly formed parallelogram was 2:1, he put the angles of the triangles at a 2:1 ratio."
That is a big assumption which I don't think you can make without some justification.
I can't understand which angles he is saying are in a 2:1 ratio? Is it c : a or b : a or something else?

Just FYI, this is how I would have done it.

<ABC = x (isosceles triangle)
<ACB = 180 - 2x (angles in a triangle add to 180)
<BCA = 2x (angles on a straight line add to 180)
<BDC =2x (isosceles triangle)

Now, considering triangle CBD, <CBD = 180 - 4x (angles in a triangle add to 180) ..... (*)

Also, considering the 3 angles lying along line AB:
x + <CBD + 60 = 180 (angles on a straight line add to 180)
Rearranging gives:
<CBD =120 - x

Previously (see (*)), we found that <CBD = 180 - 4x

Therefore 120 - x = 180 - 4x
3x = 60
x = 20.

 

[Steven G]

The top left angle in the right triangle is 2x (exterior angle rule). So the top right angle of the right triangle is also 2x.
The angle to the left of the 60 degree angle is 180-2x.
You can easily finish up from here.

 

[The Highlander]

Hi everyone! I'm a parent to some incredibly smart kids. We homeschool, and I'm having trouble finding an answer to a problem. (I'll post a photo). Any help on how to get the solution is greatly appreciated. I think we are CLOSE, but I'm coming up with something different than the answer key.

View attachment 33920
The question has two isosceles triangles connected by one of their sides (that are equal). We are given a single angle outside of the two triangles, of 60 degrees. We are asked to solve for x, which is the equal angle of a second triangle. How we are going about this is as follows:
1) We know that the left sided triangle has two angles X and the vertex angle is 180-2x.
2) we know that the vertex angle of the second triangle then has to be 120-x
3) If we label the second triangle, the opposite angles can be labeled as Y.
4) This also means that the vertex of the second triangle can also equal 120-2y.
5) The straight line across the top of the two triangles would equal 180-y, or 180-2x.
6) This is where I fail and can't figure out how to bring it home. I THOUGHT I could go y=2x at that point, or X=1/2y and then plug it back in, but I'm ripping my hair out, next to my wife and two children. One kid is trying to do bisecting lines to make right angles, the other is following in the same format I was going.

HELP!

Thanks so much in advance.

Chris

The simplest approach (as I see it) is to name the 'relevant' angles K, L & M as shown in your (modified) diagram, below (I won’t bother with any of the ‘proper’ angle (∠) or degree (°) symbols henceforth for the sake of clarity and speed of response).

​

   K = 180 - 2x (as you say)
\(\displaystyle \Rightarrow\) L = 180 - K = 180 - (180 - 2x) = 180 - 180 + 2x = 2x
\(\displaystyle \Rightarrow\) M = 180 - 4x

So, we now have:-

​

Then, considering the sum of the angles at point B, we get:-

   60+M+x = 180
\(\displaystyle \Rightarrow\) 60 + (180 - 4x) + x = 180
\(\displaystyle \Rightarrow\) 60 + 180 - 4x + x = 180
\(\displaystyle \Rightarrow\) 240 - 3x = 180
\(\displaystyle \Rightarrow\) 240 - 180 = 3x
\(\displaystyle \Rightarrow\) 3x = 60
\(\displaystyle \Rightarrow\) x = 20°

 

[Steven G]

​

  

To OP,
<L must be the sum of the two angles in the left triangle that is not <K. That is <L=2x

 

[drchrisdvm2009]

I greatly appreciate all the help and explanation. Exterior angle rule - that’s what my middle child used to get the answer, from what I’m gleaning. I’ve never heard that before (or maybe I just forgot it).

Next time I’ll label more to be clearer. Sorry about that. (First time poster). Hopefully I won’t need too much more help on proofs. (Maybe when we get to calculus).

Hope you have a great day!!