Math/Geometry problem
Moderators: Fridmarr, Worldie, Aergis, Sabindeus, PsiVen
8 posts
• Page 1 of 1
Math/Geometry problem
Hi guys.
I had a poke around the 'net but didn't find an answer on this, and I know there are a lot of mathsoriented people here, so I figured I would give it a shot.
If you have a rectangle that is 1 unit wide by 2 units tall, and you draw a diagonal from one corner to the other, what are the angles of the triangles?
Ages ago I had geometry class  around the time geometry was invented I believe  and I seem to recall some rule that in a right triangle, the angle is proportional to length the side opposite the angle. If that is the case, then one side would be one unit, and the other two units. In order to have a right triangle in which one angle is twice the size of the other, it must therefore be a 306090 right triangle.
However, if you were to reflect the triangle horizontally, you would have an isosceles triangle with a base of 2u and a height of 2u. If each lower corner of that triangle was 60 degrees, then the top angle must also be 60 degrees, which would be an equilateral triangle. This cannot be, because the height would have to equal the length of the sides. So that says the angle cannot be 60 degrees.
So where have I gone wrong?
I had a poke around the 'net but didn't find an answer on this, and I know there are a lot of mathsoriented people here, so I figured I would give it a shot.
If you have a rectangle that is 1 unit wide by 2 units tall, and you draw a diagonal from one corner to the other, what are the angles of the triangles?
Ages ago I had geometry class  around the time geometry was invented I believe  and I seem to recall some rule that in a right triangle, the angle is proportional to length the side opposite the angle. If that is the case, then one side would be one unit, and the other two units. In order to have a right triangle in which one angle is twice the size of the other, it must therefore be a 306090 right triangle.
However, if you were to reflect the triangle horizontally, you would have an isosceles triangle with a base of 2u and a height of 2u. If each lower corner of that triangle was 60 degrees, then the top angle must also be 60 degrees, which would be an equilateral triangle. This cannot be, because the height would have to equal the length of the sides. So that says the angle cannot be 60 degrees.
So where have I gone wrong?
 Koatanga
 Posts: 1985
 Joined: Mon Nov 17, 2008 12:46 pm
Re: Math/Geometry problem
This comes back to SOH CAH TOA if I recall.
http://www.mathwords.com/s/sohcahtoa.htm
http://www.mathwords.com/s/sohcahtoa.htm
Ellifain @ Khaz'Goroth does not approve of torture, save where there's experience/rep/loot involved.
 masterpoobaa
 Posts: 2230
 Joined: Thu Jul 31, 2008 5:14 pm
 Location: Brisbane, Australia, Earth, Sol, Orion Arm, Milky Way, Local Group, Virgo Supercluster, Universe.
Re: Math/Geometry problem
Imagine your triangle constructed from a bisected rectangle, so that the base is 1, height 2. Use this image and fit it to your mental picture:
In this image, b is 1, a is 2, so for your "reflected triangle" we would be looking at two angles A, one angle, 2B, all adding up to 180 degrees.
For our purposes we want to know A, and the easiest way to find it is to deal with the tangent function, since it is the opposite side divided by the adjacent. We know that tan(A) = a/b, which in this case, means that tan(A)=2. Plug arctan(2) (Arctan finds the angle given the opposite/adjacent value) into your calculator (Or Google), and it will tell you that it's equal to ~ 63.43 degrees. FYI, tan(60) is about 1.73.
The angle of B is about 26.57. So your reflection would be an isosceles with two angles of 63.43, and one of 53.14, which together gives you 180. That triangle is not equilateral, hence the answer is not 60 degrees, as you said.
In this image, b is 1, a is 2, so for your "reflected triangle" we would be looking at two angles A, one angle, 2B, all adding up to 180 degrees.
For our purposes we want to know A, and the easiest way to find it is to deal with the tangent function, since it is the opposite side divided by the adjacent. We know that tan(A) = a/b, which in this case, means that tan(A)=2. Plug arctan(2) (Arctan finds the angle given the opposite/adjacent value) into your calculator (Or Google), and it will tell you that it's equal to ~ 63.43 degrees. FYI, tan(60) is about 1.73.
The angle of B is about 26.57. So your reflection would be an isosceles with two angles of 63.43, and one of 53.14, which together gives you 180. That triangle is not equilateral, hence the answer is not 60 degrees, as you said.

Torquemada  Posts: 1678
 Joined: Mon Feb 04, 2008 3:00 am
 Location: Virginia
Re: Math/Geometry problem
Additionally, you can prove it's not an equilateral triangle by looking at the sides. Given for the original triangle that the base is 1, and height 2, then h^2=a^2 + b^2, so h=(1+4)^.5 or h is equal to the squareroot of 5. In the "reflected" triangle, the base is 2, and the other sides are 5^.5, which is not 2, thus, it is not equilateral.

Torquemada  Posts: 1678
 Joined: Mon Feb 04, 2008 3:00 am
 Location: Virginia
Re: Math/Geometry problem
Sometimes
Oscar
Herds
Cows
And
Hurls
Them
Off
Airplanes
Oscar
Herds
Cows
And
Hurls
Them
Off
Airplanes

_Chloe  Moderator
 Posts: 971
 Joined: Fri Dec 07, 2007 6:36 pm
 Location: Santa Monica, CA
Re: Math/Geometry problem
Actually I was hoping it had a tidy geometric solution involving proofs and theorems and logical goodness instead of a dull trigonometry calculation.
It should be  you have two squares divided by a diagonal, so it should have nice properties and such, but it's just a boring old nonspecial angle.
It should be  you have two squares divided by a diagonal, so it should have nice properties and such, but it's just a boring old nonspecial angle.
 Koatanga
 Posts: 1985
 Joined: Mon Nov 17, 2008 12:46 pm
Re: Math/Geometry problem
Just a quick thought about your original post... if it had resulted in an equilateral triangle, you'd be looking at a rhombus. Based on the start of a rectangle, you're guaranteed a right angle for both triangles, regardless of which corner you cut from.
Edit: If I'm misunderstanding your post, then please ignore the above.
Edit: If I'm misunderstanding your post, then please ignore the above.
"me no gay, me friends gay, me no like you call me gay, you dumb dumb" bldavis
"Here are the values that I stand for: I stand for honesty, equality, kindness, compassion, treating people the way you wanna be treated, and helping those in need. To me, those are traditional values. That’s what I stand for." Ellen Degeneres
"I'm not going to censor myself to comfort your ignorance." Jon Stewart
Horde: Clopin Dylon Sharkbait Xiaman Metria Metapriest
Alliance: Schatze Aleks Deegee Baileyi Sotanaht Danfer Shazta Rawrsalot Roobyroo
"Here are the values that I stand for: I stand for honesty, equality, kindness, compassion, treating people the way you wanna be treated, and helping those in need. To me, those are traditional values. That’s what I stand for." Ellen Degeneres
"I'm not going to censor myself to comfort your ignorance." Jon Stewart
Horde: Clopin Dylon Sharkbait Xiaman Metria Metapriest
Alliance: Schatze Aleks Deegee Baileyi Sotanaht Danfer Shazta Rawrsalot Roobyroo

Skye1013  Maintankadonor
 Posts: 3940
 Joined: Tue May 18, 2010 5:47 am
 Location: JBPHHickam, Hawaii
8 posts
• Page 1 of 1
Who is online
Users browsing this forum: No registered users and 1 guest