Trigonometry HELP rationalizing a denominator and special triangles:?

Question

Could someone post the answer to these problems and show how to do them please.... I don't have a clue about finding the missing sides.... help

http://www.mathhelpforum.com/math-help/trigonometry/24849-rationalizing-denominator.html

Answer 1

30/60/90 and 45/45/90 right triangles are very common, and it is handy to know the RATIO of the three sides of each triangle. For the 45/45/90 right triangles, the ratio is 1:1 for the two sides of the right angle, and the hypotenuse is sqrt(2) times either side. So if you recognize that you have such a triangle and know ANY side, you can find the others by remembering this ratio. The math may be a bit tricky. Try the last problem. The hypotenuse is 6. So for the other sides to have the ratio of 1 to sqrt(2), you can make a proportion. Call x the length of either side. Then
x/1 = 6/sqrt(2). To clear the denominator of the square root term, we mult BOTH numerator and denominator by sqrt(2) and now x= 6 sqrt(2) /2 or 3sqrt(2). With this you should be able to handle the other 2 45/45/90 triangles.

With the 30/60/90 triangle, the ratios are
for the side OPPOSITE the 30 degree angle (the shortest side) =1
for the side OPPOSITE the 60 degree angle (the other non-hypotenuse side)= sqrt(3)
for the hypotenuse=2.
For the three problems, this is really all you need to do them. For example, in problem 3, we have that the hypotenuse is 6. From the discussion, the "short" side is 3, and the side opposite the 60 degree angle is 3sqrt(3).