14. the shortest side across from 30 degrees is 7 and the hypotenuse is always twice the length of the shortest side so y=14 and the middle length which is across from 60 degrees is always half the hypotenuse and root 3 so x= 7[tex]\sqrt{3}[/tex]
16. The two sides (x and y) are congruent because it is an isos triangle so the sides would be half of the hypotenuse and root 2 so x&y= [tex]3\sqrt{2}[/tex]
18. This problem has both triangles in them so by using the same technique, just pretend the triangles are two separate triangles. So since it already has a root 2 in it then x= 24 and half of 24 is 12 so y= 12
