G@y GEometry help?

Question

http://img122.imageshack.us/img122/764/scan0001jh7.jpg

Find x, and y.

Answer 1

1.) For the first problem, y = 15, x = 5 √3.

The value of y follows because the downwards projection of the side labelled 10 gives you the small length of 10 cos 60 = 10*1/2 = 5, and subtracting THAT from 20 (which is in fact the correct DE-projection of that same 10 in the largest triangle) leaves you with y, which is therefore 15.

Meanwhile, the vertical line of length x = 10 sin 60 = 10 √3/2 = 5 √3.

So that's done. QED


2.) In the second problem, BC = √2 x = 16 cos 30 = 16√ 3/2 = 8√ 3. Therefore:

x = 8 √(3/2) = 9.79796... . QED

(Even shorter!)

Both very simple really. But tell me, why do you call it "Gay Geometry"? Considering that all the lines are straight, wouldn't it be more appropriate to call it "Straight Geometry"?

In any event, live long and prosper.

P.S. aspx, my students are sufficiently sophisticated to recognize projections and de-projections when they see them, such as x/sin 45 = √2 x = BC in the 2nd problem. (You used the even more sophisticated properties of both cos 30 and cos 60, without comment or proof of their values, in problem 1. How were those known? Also "By magic"?)

They certainly would regard a derivation by way of (i) proving the existence of an isosceles triangle, and (ii) Pythagoras's Theorem, as unnecessarily cumbersome, given the goal in this second problem.

Answer 2

1)
BC^2 = AB^2 + AC^2
40^2 = AB^2 + 10^2
AB^2 = 400 - 100 = 300
AB = sqrt(300)

y= AB.cos(30) = sqrt(300) * sqr(3) * 1/2 = (sqrt(3))^2 * 10 * 1/2 =30/2 = 15

x=AB.cos60 = sqrt(300) * 1/2
= 10* sqrt(3) * 1/2 = 5*sqrt(3)

2)

angle of ABC = 180 - 90 - 45 = 45
==> triangle ABC is isosceles ==> AC = AB = x

BC^2 = AC^2 + AB^2 = x^2 + x^2 = 2.x^2
BC = x.sqrt(2)

also, BC= BD.cos30 = sqrt(3) * 1/2 * 16 = 8*sqrt(3)
BC=BC ==> x.sqrt(2) = 8sqrt(3)
==> x= 8*sqrt(3)/sqrt(2) = 9.798 (approximately)

where sqrt is √

* * * * * Edit * * * * *
Dr. Spock, How did you determine BC = √2 x ? By magic? No steps. Do you teach your students this way?
* * * * * * * * * *
What i meant by "your students" are those who do not know these stuff at a certain level of education.

Isosceles and Pythagoras are quite simple and easy for students to understand and memorize. So, where i had the chance to use it, i did.

But this is not what i was pointing to. I wouldn't say a word if you just wrote

x=BC.sin45
and then BC = x/sin45
BC = sqrt(2).x

but you wrote the last line directly leaving the reader who is just learning these stuff with no idea on how to determine it.

Answer 3

In the right hand right triangle, the hypotenuse is 10, so the short side (next to the 60°) is 5, and x is 5√3. In the left right triangle, x is half of the hypotenuse, which must be 10√3, and y will be x√3 = (5√3)(√3) = 5(3) = 15.