Math help?

In a #30-60-90# special right triangle:
Side opposite to #30# degrees: #x#
Side opposite to #60# degrees: #xsqrt3#
Side opposite to #90# degrees: #2x#

So in this case we are given the side opposite to the #60# degrees, because the hypotenuse is always the opposite side of #90# degrees, and the shortest side is always opposite to the shortest angle so in this case #30# degrees:

#xsqrt3=2sqrt3#

Solve for #x# which is the shortest side:
#x=2#

Now solve for the hypotenuse:
#2x= 2*2= 4#

Now referring to the 45-45-90 triangle with the knowledge that the hypotenuse of the previous triangle was #4#, but the hypotenuse for that triangle is actually a leg of this isosceles right triangle:

An isosceles right triangle, is always a 45-45-90 degrees triangle, therefore:
Side opposite to 90 degrees: #xsqrt2#
Side opposite to 45 degrees: #x#
Side opposite to 45 degrees: #x#

Since #x# is also side opposite to #45# degrees:
#x=4#

From the right angled triangle RST we have

#sin60=(RS)/(RT)=>RT=(2sqrt3)/(sqrt3/2)=>RT=4#

From the right angled triangle RQT we have

#tan45=(RQ)/(RT)=>RQ=RT*tan45=4*1=4#

Hence #x=RQ=4#