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 ?

score 4 · Accepted Answer · answered Jan 10 '14 at 23:35

4

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

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

answered Jan 10 '14 at 23:35

Fly by Night

32,272

Thanks a lot for the clear and fast answer (both you and Mike), the link to Khan Academy and the implied tip to use proper formatting next time. – Ilija Boshkov Jan 11 '14 at 01:04
@IlijaBoshkov No problem. It was my pleasure. – Fly by Night Jan 11 '14 at 16:44

score 1 · Answer 2 · answered Jan 10 '14 at 23:31

1

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 $\sqrt{13}$.

answered Jan 10 '14 at 23:31

Mike

13,318

A little explanation of square root in a fraction

2 Answers2