Originally posted by geepamoogleOkay, geepamoogle's answer was killing me because I knew I could do better.
4) Make an triangle by mirroring the triangle along the long leg. You now have an equilateral triangle, and since the hypotenuse is one side, and the short leg is half aside, the hypotenuse must be twice the short leg, or 10 cm.
--- Original Question ---
4. In a 90 degree triangle, ABC, side A is the short leg, B the long leg, and C the hypotenuse. One of the other angles = 30 degrees. Leg A = 5 cm. Find length of the hypotenuse, without using sin/cos/tan.
--- Great Answer ---
Knowing:
1) That the circumcentre of any triangle is the point at which all three vertices are the same distance from each other (def'n), and
2) Knowing that the circumcentre for a right triangle is the mid-point of the hypotenuse (property of right triangles), and
3) Knowing (because you're not stupid) that the other angle (B) must be 60 degrees (property of triangles)...
a) A line drawn from the right angle (C) to the midpoint of the hypotenuse (M) will create two isosceles triangles - AMC and BMC.
b) BMC must be equilateral because it is give n that one of the base angles is 60 degrees, therefore the other base angle is... therefore the top angle is too...
c) Therefore half of the hypotenuse is 5 cm as all sides of the equilateral are 5 cm and the total length must therefore be 10.
This only works for this case because of the 60 degree angle. It is a more complicated solution that geep's (which I like for its simplicity), but this one is a real brain freeze.