Equilateral triangle $ABC$ has $P$ on $AB$ and $Q$ on $AC$. The triangle is folded along $PQ$ so that vertex $A$ now rests at $A'$ on side $BC$. If $BA'=1$ and $A'C=2$ then the length of the crease $PQ$ is
$\text{(A) } \frac{8}{5} \text{(B) } \frac{7}{20}\sqrt{21} \text{(C) } \frac{1+\sqrt{5}}{2} \text{(D) } \frac{13}{8} \text{(E) } \sqrt{3}$
$ABC$ has side length $3$. Let $AP=A'P=x$ and $AQ=A'Q=y$. Thus, $BP=3-x$ and $CQ=3-y$. Applying Law of Cosines on triangles $BPA'$ and $CQA'$ using the $60^{\circ}$ angles gives $x=\frac{7}{5}$ and $y=\frac{7}{4}$. Applying Law of Cosines once again on triangle $APQ$ using the $60^{\circ}$ angle gives \[PQ^2=\frac{(21)(49)}{400}\] so \[PQ=\frac{7}{20}\sqrt{21}\] The correct answer is $\fbox{(B)}$.