1986 AMC 12 Problem 21

Let the circle’s radius be $r=AC$ and compare a sector with triangle $ABC$

Equality of the two shaded pieces means the whole triangle has twice the sector’s area

Let $r=AC.$ The upper shaded sector has area $\frac{\theta r^2}{2}.$ The lower shaded region is triangle $ABC$ with an equal sector removed. Thus the two shaded regions are equal exactly when $$2\left(\frac{\theta r^2}{2}\right) =\frac12r\cdot AB.$$ This is equivalent to $\frac{AB}{r}=2\theta.$ Since $AB$ is tangent at $A,$ triangle $ABC$ is right at $A,$ and $\frac{AB}{r}=\tan\theta.$ Therefore the condition is $\tan\theta=2\theta.$

Thus the correct answer is B.