
Algebra problem
Hello! I have a problem with an angle. I just cant calculate it. But I'm pretty sure that this is possible. What do you say!? I would be more than happy if some of you guys could see how to do this. The problem is depicted on this adress.
http://engman.bravehost.com/Jobb/algebra.JPG
/Gustav from Sweden
Re: Algebra problem
I have some questions about the diagram. With the measurement of 1.5m, is it meant to stop at the point where the circle touches the line, or does it stop where the diagram says it stops? I ask because as the diagram is, that measurement is currently useless.
Also, the radius of the circle is shown as 0.333m. There's nothing wrong with that, although it makes calculations more horrible, so I was wondering/hoping if that measurement is meant to be 1/3m?
Why did the vector cross the road? It wanted to be normal.
Re: Algebra problem
For the moment, I'm assuming that the diagram is right, so I'm not using the 1.5m measurement and and I'm taking the radius as 0.333m.
My method starts off with drawing the diagram on a graph. The centre of your angle is (0,0), the centre if the circle is (1,0.333) and the point where the circle touches the top line is (p,q), where p and q are unknowns.
A way of working out the angle is to find the gradient of the top line. As the bottom line has a gradient of 0, the angle would be the inverse tan of the top line's gradient. We know that this is q/p, so these two values are what we must find. These are two values, so we need two equations involving these values.
One of these equations can be found by finding the gradient of the line between the point where the circle touches the top line and the circle's centre. The circle's centre is (1,0.333) and the touchpoint is (p,q), so we know the gradient is (0.333q)/(1p). However, as this line is a radius of the circle and the top line is a tangent to the circle, these two lines must be perpendicular to each other. This means that one of their gradients is the negative inverse of the other one, meaning the gradient can also be represented as p/q. First equation: (0.333q)/(1p)=p/q
The second equation can be found by knowing that (p,q) is touching the circle and so must be 0.333m away from its centre. Making a rightangled triangle with the radius as the hypotenuse, we can see that the other lengths will be (q0.333) and (1p) Hence, using Pythagoras: Second equation: sqrt((q0.333)²+(1p)²)=0.333
Using these two equations, you should be able to work out p and q. Ooh, I just remembered, I have to go over there to, um, do something...
Why did the vector cross the road? It wanted to be normal.
Re: Algebra problem
Heyy Mathsyperson! That is such a cool answer, it's a real pleasure to read it! But, Im convinced that the problem that this algebra subproblem is conected with, is not ment to be this advanced... so you said that the 1.5m measurement is useless. It seems so to me to... but it should be involved somehow. So, you are absolute positive that that measurement cannot unlock an easier solution to this?
Re: Algebra problem
It seems strange to me too. I think it's just a mistake on the diagram, but at the moment one end of the measurement is at the corner of your wanted angle, and the other end stops at nowhere significant. If it stopped where the circle touches the line, then it could be used to get an easier solution, but at the moment it is just sitting there looking pretty.
Why did the vector cross the road? It wanted to be normal.
 MathsIsFun
 Administrator
Re: Algebra problem
Another way to look at it is:
Solve a triangle that goes straight to the centre of the circle. It is a rightangled triangle, with two sides of 1m and 0.333m. That angle would be tan1(0.333/1), then double that angle for the result (by symmetry).
I figure it is 36.8°
(Gustav  are you drawing those yourself, or just adding "I want his angle so bad"? If you are, then Good Drafting!)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
Re: Algebra problem
I am extremely annoyed by the easiness of your method and hence the complete pointlessness of mine. Having said that, well done!
Why did the vector cross the road? It wanted to be normal.
 MathsIsFun
 Administrator
Re: Algebra problem
Only because I saw the symmetry  more luck than good management!
BTW, I recently saw a page with 50+ ways of proving Pythagoras, so isn't maths grand?
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
Re: Algebra problem
My favourite is the one with the square made up of a slightly smaller square and 4 rightangled triangles. It's easily the easiest one to understand that I've found so far.
Why did the vector cross the road? It wanted to be normal.
Re: Algebra problem
You have both been of great help, and I'm so greatful. To MathIsFun: That was the solution I knew was there. Mathsyperson: I'm impressed! And regarding the "i want this angle so bad"; It is written by me, and have been more than true for the last 72 hours. But not any more... hihi... it's like a burden has been lifted from my shoulders, and the salvator: "www.mathisfun.com". You guys rock!
 MathsIsFun
 Administrator
Re: Algebra problem
* collective bow with mathsyperson shoved forwards *
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
 ganesh
 Moderator
Re: Algebra problem
mathsyperson wrote:My favourite is the one with the square made up of a slightly smaller square and 4 rightangled triangles.
And thats the only one I remember!
Character is who you are when no one is looking.
Re: Algebra problem
hi i dont now a clue on algrbra can you tell me some stuff
Re: Algebra problem
hi i dont now a clue on algrbra can you tell me some stuff
 ganesh
 Moderator
Re: Algebra problem
Algebra is using of variables like x,y,z etc to solve mathematical problems. For example, if you have this problem #1 When 53 is added to a number, you get 70. What is the number? The solution would start this way. Let x be the number. Therefore, 53 + x = 70 x = 70  53 = 17
#2 25 added to the double of a number is 95. What is the number? Solution: Let x be the number. Therefore, 25 + 2x = 95 2x = 95  25 = 70 x = 70/2 = 35
Character is who you are when no one is looking.
 MathsIsFun
 Administrator
Re: Algebra problem
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
Re: Algebra problem
If you are in harder algebra try a bit prealgebra again
 MathsIsFun
 Administrator
Re: Algebra problem
Yes, it is always worth while going back and revising the basics. I do it myself.
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
Re: Algebra problem
A postulate can be derived from the original conversation:
sin θ / (1 + cos θ) = tan ( θ / 2 )
igloo myrtilles fourmis
 MathsIsFun
 Administrator
Re: Algebra problem
sin θ / (1 + cos θ) = tan ( θ / 2 ) For 30°: (1/2)/(1+√(3)/2) = tan 15° ==> 0.267949 = 0.267949
Yes! (at least for 30°)
"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  Leon M. Lederman
