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...

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.

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

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.

* collective bow with mathsyperson shoved forwards *

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

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 

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

A postulate can be derived from the original conversation:

sin θ / (1 + cos θ) = tan ( θ / 2 )