I have a circle that is divided into 4 quadrants with a vertical and a horizontal axis.
The center of the circle (where the axes cross) is point b.
The top of the vertical axis is point d.
On the horizontal axis, to the left of b is point a.
A line is drawn from a diagonally up and to the right, across the vertical axis (at point c) and meets the top right of the circle at e.
Line be and point f (the right end of the horizontal axis) have been added to help with the solution.
The radius of the circle is 30 and the x position of point a is -.5 from b, taking b to be (0, 0) (that is, a is to the left of b by .5)
How would we go about calculating $ \angle eab$ so that $\triangle cab$ = slanty sector dce?
With trial and error in a graphing program, I found the answer to be 88.09075925431...
But how would one go about solving this in a general method?
Here is an outline of the solution:
Let $\theta = \angle dbe$, in radians. Then we get the area of the sector to be $900\pi\cdot\frac{\theta}{2\pi} = 450\theta$. And the area of $\triangle abe$ is $\frac12 (ab)(be)\sin(\frac\pi 2+\theta) = \frac{15}2\cos\theta$. Setting these equal, we get the equation $$60\theta = \cos\theta.$$ We can solve this numerically (e.g., Newton's method) or by using the second-order Taylor polynomial of $\cos$ and the quadratic formula. Either way, we get $\theta\approx 0.0166644$.
Now you can use standard trigonometry (e.g., law of cosines and law of sines) to find your angle $eab$ to be approximately $1.53747$ radians, or $88.0905^\circ$.