Question 1066460

Hi! 

So, in my geometry class we are learning about special right triangles. (30,60,90 and 45,45,90). I recently took a quiz over the information and I was stuck on one problem. There is a 30,60,90 triangle, with the long leg as 10 square root three, and the short leg as X. The hypotenuse is y. The problem wanted you to solve for x and y. I tried using the rule of: short leg= x, long leg=x square root three, and the hypotenuse= 2x. Using that rule I got x as 10 and y as 20. Unfortunately this wasn't the right answer, and I was wondering if you could work through the problem, step by step to help me understand how to use the 30,60,90 rule and how it applies to this problem. 

Thank you so much!! 
<pre>In a 30-60-90 special triangle, if {{{matrix(1,4, Longer, leg, "=", 10sqrt(3))}}}, then:

Shorter leg = {{{matrix(1,4, Longer, leg, "*", sqrt(3))/3)}}}. In this case, we get: Shorter leg, or {{{highlight_green(matrix(1,9, X, "=", (10sqrt(3) * sqrt(3))/3, "=", 10(3)/3, "=", 10cross((3))/cross(3), "=", highlight(matrix(1,2, 10, units))))}}}
Hypotenuse, or {{{highlight_green(matrix(1,10, y, "=", 2, "*", Shorter, "leg,", or, 2(10), "=", highlight(matrix(1,2, 20, units))))}}}
Your answers appear to be CORRECT!