An arcPQ of a circle of radius $r$ is created with a central angle $($in radians$)$ as shown in the figure to the right. Assume the center of the circle is the origin of the Cartesian coordinate plane $($so the point $R$ lies on the $x-$axis$).$

$($a$)$ Find the equations of lines tangent to the circle at $P$ and $Q($you should have three variables in your answer: $x,$ the unspecified angle $,$ and the unspecified radius $r).$

$($b$)$ These two tangent lines intersect at a point $R.$ Find it $($your answer will also include $$ and $r).$

$($c$)P,Q,$ and $R$ form a triangle as shown in the figure. What is the formula for the area of this triangle, in terms of $$ and $r?$

$($d$)$ Recall that the formula for the area of a segment $($piece of circle cut off from the rest by a line segment $PQ$ is $A()=\frac{r^2}{2}(-sin).$ Let $B()$ be the area between the line segments $PR,QR,$ and the arc $PQ.$ Calculate $\underset{0^{+}}{lim}\frac{A()}{B()}.$

$($e$)$ Why do you think the ratio in Part D approached the number you got? $($Don$\text{'}$t just say "because that's what we got." Try to come up with an explanation that doesn't use your calculations.$)$