A5. What is the perimeter of a right triangle if one interior angle is $60^{\circ}$ and the lengths of the legs differ by 2 units?
(A) $4 \sqrt{3}+6$
(B) $4 \sqrt{3}-6$
(C) $4 \sqrt{3}+2$
(D) $4 \sqrt{3}+4$
(E) 6
A5. We note that the lengths of the legs are $x$ and $x+2$. We use the trigonometric function tan $60^{\circ}=\sqrt{3}$. We set up the equation $\sqrt{3}=\frac{x+2}{x}$. We multiply the equation by the common denominator $x$. We get the equation $\sqrt{3} x=x+2$, rearrange it to $x(\sqrt{3}-1)=2$. We express $x=\frac{2}{\sqrt{3}-1}$, rationalize with $\sqrt{3}+1$ and get $x=\sqrt{3}+1$. We then calculate the perimeter of the triangle $o=a+b+c=$ $\sqrt{3}+1+\sqrt{3}+3+2 \sqrt{3}+2=4 \sqrt{3}+6$.