Equilateral triangle and perpendicular line

I didn't quite know what to put as title, so I'm sorry if the title doesn't give a clue of what is to be found (Itwasntme)

Well, here's the question:

Qu: The points S and T are located on the respective sides PQ and PR of an equilateral triangle PQR so that ST = TR and ST is perpendicular to PQ. Given that the length of QR is 1, the length of ST is

A. $\displaystyle \frac{1}{2}$

B. $\displaystyle 2-\sqrt{3}$

C. $\displaystyle 2\sqrt{3} - 3$

D. $\displaystyle 2(2-\sqrt{3})$

E. $\displaystyle \frac{1}{3} \sqrt{3}$

I proceeded with finding the angles, angle STR being 150 degrees and angle TSR = angle TRS = 15 degrees.

Then, I found the length of QS, since QRS is a right angle with one of the angles being 45 degrees. QS = $\displaystyle \frac{1}{\sqrt{2}}$ = SR.

Then, fron the sine rule, I know that:

$\displaystyle \frac{L}{sin(15)} = \frac{1}{\sqrt{2} sin(150)}$

So,

$\displaystyle L = \frac{2}{\sqrt{2}sin(15)}$

Now, I don't know the exact form on sin(15) and I don't see any square root of 2 in any of the choices. Is there a quicker/simpler way to solve this?

Thanks!