Algebra problem

GurraTedden · 2005-07-20 04:06:01

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

mathsyperson · 2005-07-20 06:05:11

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?

mathsyperson · 2005-07-20 06:46:42

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 touch-point is (p,q), so we know the gradient is (0.333-q)/(1-p).
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: (0.333-q)/(1-p)=-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 right-angled triangle with the radius as the hypotenuse, we can see that the other lengths will be (q-0.333) and (1-p)
Hence, using Pythagoras:
Second equation: sqrt((q-0.333)²+(1-p)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...

GurraTedden · 2005-07-20 07:21:58

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?

mathsyperson · 2005-07-20 08:59:16

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.

MathsIsFun · 2005-07-20 10:31:13

Another way to look at it is:

Solve a triangle that goes straight to the centre of the circle. It is a right-angled triangle, with two sides of 1m and 0.333m. That angle would be tan-1(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!)

mathsyperson · 2005-07-20 18:12:24

I am extremely annoyed by the easiness of your method and hence the complete pointlessness of mine.
Having said that, well done!

MathsIsFun · 2005-07-20 18:22:22

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?

mathsyperson · 2005-07-20 18:32:14

My favourite is the one with the square made up of a slightly smaller square and 4 right-angled triangles. It's easily the easiest one to understand that I've found so far.

GurraTedden · 2005-07-20 20:20:10

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 · 2005-07-20 20:54:27

* collective bow with mathsyperson shoved forwards *

Jai Ganesh · 2005-07-20 21:17:54

mathsyperson wrote:

My favourite is the one with the square made up of a slightly smaller square and 4 right-angled triangles.

And thats the only one I remember!

Thushika · 2005-07-22 17:50:40

hi i dont now a clue on algrbra can you tell me some stuff

se · 2005-07-22 17:51:55

hi i dont now a clue on algrbra can you tell me some stuff

Jai Ganesh · 2005-07-22 17:59:20

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

MathsIsFun · 2005-07-22 18:38:08

Have a try solving some yourself, here: http://www.mathsisfun.com/worksheets/pr … 4&ID=20709

skatergirl · 2005-08-05 04:52:09

If you are in harder algebra try a bit pre-algebra again

MathsIsFun · 2005-08-05 11:53:25

Yes, it is always worth while going back and revising the basics. I do it myself.

John E. Franklin · 2005-09-05 06:09:19

A postulate can be derived from the original conversation:

sin θ / (1 + cos θ) = tan ( θ / 2 )

MathsIsFun · 2005-09-05 10:06:36

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°)

Math Is Fun Forum

#1 2005-07-20 04:06:01

Algebra problem

#2 2005-07-20 06:05:11

Re: Algebra problem

#3 2005-07-20 06:46:42

Re: Algebra problem

#4 2005-07-20 07:21:58

Re: Algebra problem

#5 2005-07-20 08:59:16

Re: Algebra problem

#6 2005-07-20 10:31:13

Re: Algebra problem

#7 2005-07-20 18:12:24

Re: Algebra problem

#8 2005-07-20 18:22:22

Re: Algebra problem

#9 2005-07-20 18:32:14

Re: Algebra problem

#10 2005-07-20 20:20:10

Re: Algebra problem

#11 2005-07-20 20:54:27

Re: Algebra problem

#12 2005-07-20 21:17:54

Re: Algebra problem

#13 2005-07-22 17:50:40

Re: Algebra problem

#14 2005-07-22 17:51:55

Re: Algebra problem

#15 2005-07-22 17:59:20

Re: Algebra problem

#16 2005-07-22 18:38:08

Re: Algebra problem

#17 2005-08-05 04:52:09

Re: Algebra problem

#18 2005-08-05 11:53:25

Re: Algebra problem

#19 2005-09-05 06:09:19

Re: Algebra problem

#20 2005-09-05 10:06:36

Re: Algebra problem

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