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Triangle Problem from /b/

Name: Anonymous 2008-03-30 23:18

http://i29.tinypic.com/i40e8n.jpg

Method for obtaining a unique solution:

Assign an arbitrary length to the equal sides.  L.
Assume the bottom left corner is at the origin and derive the equations of the lines to find the endpoints of DE.
Turn DE and AE into vectors and take the dot product.

Line AD:
y = tan(80)*x

Line BD:
(2LSin10, 0)
m = -tan(60)
y = -tan60 (x - 2LSin10)

Point D:
tan(80)x = -tan60 (x - 2LSin10)
(tan80 + tan60)x = 2Lsin(10)tan(60)
x = 2Lsin(10)Tan(60) / (tan80 + tan60)
y = 2Lsin(10)Tan(60)Tan(80) / (tan80 + tan60)

Line AE:
y = tan(70)*x

Line BE:
(2LSin10, 0)
m = -tan(80)
y = -tan(80) (x - 2LSin10)

Point E:
tan(70)*x = -tan(80) (x - 2LSin10)
(tan70 + tan80)*x = 2LSin(10)Tan(80)
x = 2LSin(10)Tan(80) / (tan70 + tan80)
y = 2LSin(10)Tan(80)Tan(70) / (tan70 + tan80)

Vector DE:
x: 2Lsin(10)Tan(60) / (tan80 + tan60) - 2LSin(10)Tan(80) / (tan70 + tan80)
y: 2Lsin(10)Tan(60)Tan(80) / (tan80 + tan60) - 2LSin(10)Tan(80)Tan(70) / (tan70 + tan80)

Each term has 2Lsin(10) which can be factored out, Leaving:
<.43969,.52400>

Vector AE:
x: 2LSin(10)Tan(80) / (tan70 + tan80) - 0
y: 2LSin(10)Tan(80)Tan(70) / (tan70 + tan80) - 0

Again factoring out 2Lsin(10) leaves us with:
<.63736,1.8508>

DE (dot) AE = |DE| * |AE| * Cos(x)
1.2500 = 1.3389 * Cos(x)
Cos(x) = .93354
x = 21

Name: Anonymous 2008-03-31 2:09

>>1
Entire point was to do it with basic geometry. It's trivial with trig.

Name: Anonymous 2008-03-31 2:33

>>2
Yeah, except it's not.

Name: Anonymous 2008-03-31 12:29

>>3
It is if you aren't a fucking high school student. Incidentally, I didn't even check before posting >>2, but your answer is wrong.

Name: Anonymous 2008-03-31 18:26

>>4
I love how people keep claiming it's trivial, yet in three threads on the subject no one has easily solved it.

Name: CSharp !FFI4Mmahuk 2008-03-31 20:34

>>5
That's because he's a troll douchebag. If you do it from angles alone, you end up getting an unsolvable system. You either have to do it with trig or construct additional lines; one of the ways I saw was by inscribing certain angles within a circle.

Name: Anonymous 2008-03-31 21:09

>>6
I didn't say you couldn't construct lines. Using trig, however, is not allowed in the regular statement of the problem. The source of the problem (afaik) is http://thinkzone.wlonk.com/MathFun/Triangle.htm

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