How to Solve the Circle and Triangle Problem from the Oxford Maths Assignment?

[gangsta316]

http://www.maths.ox.ac.uk/filemanager/active?fid=2929
Question 4 on page 12.

Could someone give me some hints on how to do this? I have no idea how. All I know is that we could draw a line from the center of the circle to Q, which I think would be perpendicular to that tangent. Also if we draw a horizontal line from the center of the circle to the tangent, the angle between them will be theta.

 

[HallsofIvy]

So you have a circle with center at (1,1), radius 1, and a line tangent to it at Q making angle \theta with the x-axis. The angle \theta, as shown, is actually with the "backward pointing" x-axis. The angle with positive x-axis, the one used in defining the slope of a line is \pi- \theta. Since [/itex]tan(\pi- \theta)= - tan(\theta)[/itex], the slope of the tangent line is -tan(\theta) (oops- that was one of the things you were supposed to prove!).
Of course, the radius of the circle, from (1, 1) to Q, is perpendicular to the tangent line. Let Q= (x,y). Then the line from (1,1) to Q has slope (x-1)/(y-1) which must be the negative reciprocal of -tan(\theta) which is cot(\theta). The equations
\frac{y-1}{x-1}= \frac{cos(\theta)}{sin(\theta)}
and
x^2+ y^2= 1
should give you what you need.

 

[gangsta316]

Thanks. I managed to do that part. Are the co-ordinates of P

P (1 + \csc \theta + \cot \theta, 0)?

I can't do part (ii). For the explanation I'm unsure. All I know is that angle PRO (in the B area) is \frac{\pi}{2} - \theta.

I got A(\frac{\pi}{4}) = 1 + \sqrt2 -\frac{3\pi}{8}. Is that correct?