Can /sci/ do basic geometry?

Forgot to mention: post your answers in radical form. No decimals.

>>8645560
You're out of luck. There's a branch of mathematics called Galois theory that proves that solutions to this kind of problem can't be expressed in terms of radicals.

>>8645557
No

>>8645557

We are presented with an "overall" figure, a right triangle, with certain other information. That right triangle also has its altitude described and its other leg named. We are to find those two lengths.

First, I remark that the "large" triangle, not bothering to describe its points as my meaning should and will be plain in the exposition, is a "30-60-90" triangle, and consequently its sides always stand to one another in the proportions 1 : √3 : 2, which is determined in this case since we have all angles, and thier positions as it were. as 1 stands to √3, so must 4√3 stand in this case to y, so that I say that y is equal to 12.

The hypotenuse of the large triangle is a line segment, cut by another line segment its altitude. So that I may find the supplementary angle of the given right angle, which is again another right angle, so that they are two right angles on that line segment. But this entails two smaller triangles, each having given angles of thirty and sixty degrees respectively, so that I may remark that each item has for its remaining angles now sixty and thirty degrees. I furthermore note that these two are yet consistent. The result is that the large 30-60-90 triangle is cut into two smaller 30-60-90 triangles, which I may now unambiguously refer to as the small, the middle, and the large triangles.

It remains to find x the long leg of the small triangle, or the short leg of the middle triangle. To complete the diagram, we also propose to find the hypotenuse of the original large triangle, and its entailed bifurcation into legs for the smaller. Scalar knowledge of all 30-60-90 triangles together with what has been said above allows us to say that the short leg of the small triangle is 2√3, while the long leg of the middle triangle is 6√3, whose sum is 8√3, which is of course just again the hypotenuse of the original large triangle. It remains to find x, which is six.

Quod Erat Demonstrandum.

>>8645557
Give me numbers to crunch and I'll crunch the fuck out of them. Give me geometry and I'll shove a ruler up your ass

As a corollary of this proof, notice that for a given 30-60-90 triangle, the lengths of its altitude perpendicular to its hypotenuse, its short leg, its long leg, and its hypotenuse stand to one another in the relationship 6 : 4√3 : 12 : 8√3

>>8645557
sin 60 = sqrt(3)/2 = x/4sqrt(3)
x = 6

sin 30 = 1/2 = 6/y
y = 12

reported

>>8645557
x is 6
y is 12

30: 60: 90 relationship
1x : (root3)x : 2x

>>8645557
as the square root of 3 doesn't exist, this problem is unsolvable

>>8647026
Yet the approximation can still be useful. A house structure need not be perfect, merely close and fully designed before building commences; no measuring at the end.

>>8645571
>theory
>proves

>>8645557
*trig

>>8647334
spotted the retard who doesn't know the difference between a theory and an idea

>>8645557
6 and 12.
3d trig is far more interesting.

x = 4 x sqrt3
y = 6 x 2

>>8645560
How about you go fuck yourself?

>>8645557
>basic geometry
>trigonometry diagram