A little explanation of square root in a fraction

Question

I was on KhanAcademy when I ran into a problem involving a sine of a triangle, this was the solution:

$$\frac{9}{3\sqrt{13}}$$

(9 being the length of the opposite and $3\sqrt{13}$ being the length of the hypotenuse)

And in the given answers there was this:

$$\frac{3\sqrt{13}}{13}$$

The hints gave me this:

$$\sin(\angle ABC) = \frac{9}{3\sqrt{13}} = \frac{3\sqrt{13}}{13}$$

Can anyone explain how the first went into the second one ?

Answer 1

First we cancel down $\tfrac{9}{3}$ to give $3$ and then we rationalise the denominator:

$$\frac{9}{3\sqrt{13}} = \frac{3}{\sqrt{13}} = \frac{3}{\sqrt{13}} \times 1 = \frac{3}{\sqrt{13}}\times \frac{\sqrt{13}}{\sqrt{13}}=\frac{3\sqrt{13}}{(\sqrt{13})^2}=\frac{3\sqrt{13}}{13}$$

Answer 2

First, numerator and denominator were divided by $3$. Then, it's standard to eliminate square roots in the denominator, which in this case meant simply multiplying numerator and denominator by $\sqrt{13}$.