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[quote=mathsyperson]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. 
[b]First equation: (0.333-q)/(1-p)=-p/q[/b]

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:
[b]Second equation: sqrt((q-0.333)²+(1-p)²)=0.333[/b]

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...[/quote]

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Topic review (newest first)

MathsIsFun: 2005-09-06 08:06:36

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°)

John E. Franklin: 2005-09-06 04:09:19

A postulate can be derived from the original conversation:

sin θ / (1 + cos θ) = tan ( θ / 2 )

MathsIsFun: 2005-08-06 09:53:25

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

skatergirl: 2005-08-06 02:52:09

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

MathsIsFun: 2005-07-23 16:38:08

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

ganesh: 2005-07-23 15:59:20

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

se: 2005-07-23 15:51:55

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

Thushika: 2005-07-23 15:50:40

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

ganesh: 2005-07-21 19:17:54

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!

MathsIsFun: 2005-07-21 18:54:27

* collective bow with mathsyperson shoved forwards *

GurraTedden: 2005-07-21 18:20:10

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!

mathsyperson: 2005-07-21 16:32:14

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.

MathsIsFun: 2005-07-21 16:22:22

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: 2005-07-21 16:12:24

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

MathsIsFun: 2005-07-21 08:31:13

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!)

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