Answer
$$\lim\limits_{\theta\to0^+}\frac{A(\theta)}{B(\theta)}=0$$
Work Step by Step
1) First, find the area of the triangle $B(\theta)$
Here, we know the length of 2 triangle sides $(\overline{RP}=\overline{RQ}=10cm)$ and the included angle $(\angle R=\theta)$. Therefore, to find the area of the triangle, we would use this formula $$A=\frac{1}{2}ab\sin\angle C$$
$A$: the area of the triangle
$a$ and $b$: the length of 2 sides of the triangle
$\angle C$: the included angle of 2 sides mentioned above
That means, here in $\triangle PRQ$, its area would be $$B(\theta)=\frac{1}{2}\overline{RQ}\overline{RP}\sin\angle R$$ $$B(\theta)=\frac{1}{2}\times10\times10\times\sin\theta$$ $$B(\theta)=50\sin\theta (cm^2)$$
2) Find the area of semicircle $A(\theta)$
To find the area of the semicircle, we need to calculate $\overline{PQ}$. We can do that from the information given in $\triangle PRQ$
$\triangle PRQ$, first, is not a right triangle. Then, we already know the length of 2 sides and the included angle of these 2 sides. Therefore, the most appropriate way to find the length of the remaining side is to use Law of Cosine. In detail, $$c^2=a^2+b^2-2ab\cos\angle C$$
$a$, $b$ and $c$: the length of 3 sides of the triangle
$\angle C$: the included angle of 2 sides with the length $a$ and $b$ respectively
Applying to $\triangle PRQ$: $$\overline{PQ}^2=\overline{RQ}^2+\overline{RP}^2-2\overline{RQ}\overline{RP}\cos\angle C$$ $$\overline{PQ}^2=10^2+10^2-2\times10\times10\times\cos\theta$$ $$\overline{PQ}^2=200-200\cos\theta$$ $$\overline{PQ}^2=100\times2(1-\cos\theta)$$ $$\overline{PQ}=10\sqrt{2(1-\cos\theta)}$$
$PQ$ is the diameter of the semicircle. Therefore, the area of the semicircle would be $$A(\theta)=\frac{\pi(\overline{PQ}/2)^2}{2}$$ $$A(\theta)=\frac{\pi(5\sqrt{2(1-\cos\theta)})^2}{2}$$ $$A(\theta)=\frac{\pi\times25\times2(1-\cos\theta)}{2}$$ $$A(\theta)=25\pi(1-\cos\theta)(cm^2)$$
3) Calculate $\lim\limits_{\theta\to0^+}\frac{A(\theta)}{B(\theta)}$ $$\lim\limits_{\theta\to0^+}\frac{A(\theta)}{B(\theta)}=\lim\limits_{\theta\to0^+}\frac{25\pi(1-\cos\theta)}{50\sin\theta}=\lim\limits_{\theta\to0^+}\frac{-\pi(\cos\theta-1)}{2\sin\theta}$$ $$=\frac{-\pi}{2}\lim\limits_{\theta\to0^+}\frac{\cos\theta-1}{\sin\theta}$$
Divide both numerator and denominator by $\theta$ $$=\frac{-\pi}{2}\lim\limits_{\theta\to0^+}\frac{\frac{\cos\theta-1}{\theta}}{\frac{\sin\theta}{\theta}}$$ $$=-\frac{\pi}{2}\times\frac{0}{1}=0$$
