Math/Geometry problem

Anything, including off-topic posts

Moderators: Fridmarr, Worldie, Aergis, Sabindeus, PsiVen

Math/Geometry problem

Postby Koatanga » Mon Aug 20, 2012 5:59 pm

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 maths-oriented 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 30-60-90 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?
Retired. Koatanga, Shapely, Sultry, Doominatrix of Greenstone - Dath'Remar
Koatanga
 
Posts: 2015
Joined: Mon Nov 17, 2008 12:46 pm

Re: Math/Geometry problem

Postby masterpoobaa » Mon Aug 20, 2012 6:16 pm

This comes back to SOH CAH TOA if I recall.
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

Postby Torquemada » Mon Aug 20, 2012 6:19 pm

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:

Image

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.
User avatar
Torquemada
 
Posts: 1679
Joined: Mon Feb 04, 2008 3:00 am
Location: Virginia

Re: Math/Geometry problem

Postby Torquemada » Mon Aug 20, 2012 6:26 pm

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.
User avatar
Torquemada
 
Posts: 1679
Joined: Mon Feb 04, 2008 3:00 am
Location: Virginia

Re: Math/Geometry problem

Postby Koatanga » Mon Aug 20, 2012 6:27 pm

Thanks all.
Retired. Koatanga, Shapely, Sultry, Doominatrix of Greenstone - Dath'Remar
Koatanga
 
Posts: 2015
Joined: Mon Nov 17, 2008 12:46 pm

Re: Math/Geometry problem

Postby _Chloe » Tue Aug 21, 2012 6:53 pm

Sometimes
Oscar
Herds
Cows
And
Hurls
Them
Off
Airplanes
User avatar
_Chloe
Moderator
 
Posts: 971
Joined: Fri Dec 07, 2007 6:36 pm
Location: Santa Monica, CA

Re: Math/Geometry problem

Postby Koatanga » Tue Aug 21, 2012 7:18 pm

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 non-special angle.
Retired. Koatanga, Shapely, Sultry, Doominatrix of Greenstone - Dath'Remar
Koatanga
 
Posts: 2015
Joined: Mon Nov 17, 2008 12:46 pm

Re: Math/Geometry problem

Postby Skye1013 » Tue Aug 21, 2012 8:22 pm

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.
"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
User avatar
Skye1013
Maintankadonor
 
Posts: 3965
Joined: Tue May 18, 2010 5:47 am
Location: JBPH-Hickam, Hawaii


Return to General

Who is online

Users browsing this forum: Exabot [Bot], SteveL and 1 guest

Who is online

In total there are 3 users online :: 2 registered, 0 hidden and 1 guest (based on users active over the past 5 minutes)
Most users ever online was 380 on Tue Oct 14, 2008 6:28 pm

Users browsing this forum: Exabot [Bot], SteveL and 1 guest