# Geometry Question 2



## nitrocan (Apr 23, 2009)

ABC is a triangle.

Angles:
DAE = 20
EAB = 60
ABD = 50
DBE = 30

DEA = ?


----------



## Sa967St (Apr 23, 2009)

I figured it out  lemme write the answer now

EDIT:

spoiler below

(make the point where -EA and -BD cross 'F')

Angles:
AFB = 180-60-50 =70
DFE = AFB = 70
BFE = 180-70 = 110
DFA = BFE = 110
ADB = 180-20-110 = 50
BFE = 180-30-110 = 40

let: 
angle DEC = w
angle AED = x
angle EDC = y
angle BDE = z

y+z = 180-50 = 130
x+w = 180-40 = 140
x+z = 180-70 = 110
y+w = 180+180-110-50-40 = 160

...
x=30


----------



## JohnnyA (Apr 23, 2009)

This was my original blabble:

Call the place where the lines cross X.

AXB = 70
DXE = 70
BXE = 110
BEX = 30
CDB = 130

ADB = 40
ACB = 20

Now make a horizontal line from E across to a point above D. Call the point where this line meets DC: Y. ABC is isosceles therefore 80 and CEY = 80

DYE = 100


BUT

Wait ... in a quadralateral sums of opposite angles are equal. So, DEB + DAB = ADE + ABE.

?+120 = (50+(180-70-?)) + 80
?+120 = (50+110-?)+80
?+120 = 240-?
2?= 120
? = 60

Is this right, or am I totally wrong...


----------



## amostay2004 (Apr 23, 2009)

I calculated and got 20. Too lazy to write the solution =p I haven't touched geometry for years and I'm having an exam in 3 hours so I'm just taking a break

EDIT: checked my calculations. I'm most probably wrong. Just ignore me =p


----------



## Sa967St (Apr 24, 2009)

JohnnyA said:


> Wait ... *in a quadralateral, sums of opposite angles are equal*. So, DEB + DAB = ADE + ABE.
> 
> ?+120 = (50+(180-70-?)) + 80
> ?+120 = (50+110-?)+80
> ...



I don't think that's correct.


----------



## Vulosity (Apr 24, 2009)

Sa967St said:


> JohnnyA said:
> 
> 
> > Wait ... *in a quadralateral, sums of opposite angles are equal*. So, DEB + DAB = ADE + ABE.
> ...



True, only if the bases are parallel and crossed by a transversal, which in this case, it's false.


----------



## nitrocan (Apr 24, 2009)

I had such a different solution to this. I never thought of something like that. The right solution is Sa967St's by the way.

I might come up with some more stuff later since it's really getting late here.


----------



## amostay2004 (Apr 24, 2009)

Yea...I was calculating it the same way as Sa967S, only I screwed up some parts


----------



## keemy (Apr 24, 2009)

i solved it analytically for the lulz

ArcSin[Sin[ 20 Degree] (((((2 Tan[50 Degree])/(Tan[80 Degree] +Tan[50 Degree]))^2 + ((2 Tan[50 Degree] Tan[ 80 Degree])/(Tan[80 Degree] + Tan[50 Degree]))^2)^(1/ 2) )/((((2 Tan[80 Degree])/(Tan[60 Degree] + Tan[80 Degree])) - ((2 Tan[50 Degree])/(Tan[80 Degree] + Tan[50 Degree])))^2 + (((2 Tan[60 Degree] Tan[ 80 Degree])/(Tan[80 Degree] + Tan[60 Degree])) - ((2 Tan[50 Degree] Tan[80 Degree])/(Tan[80 Degree] + Tan[50 Degree])))^2)^(1/2) )]

which i then typed into a calculator a few times till i got a reasonable looking answer (it's hard to type all that!!) and got 30 degrees XD.


----------



## qqwref (Apr 24, 2009)

Sa967St said:


> y+z = 180-50 = 130
> x+w = 180-40 = 140
> x+z = 180-70 = 110
> y+w = 180+180-110-50-40 = 160
> ...



Wait, what? It's true that x=30 but you can't solve those four equations to get x (another solution is w=100, x=40, y=60, z=70). So what did you do?


----------



## keemy (Apr 24, 2009)

qqwref said:


> Sa967St said:
> 
> 
> > y+z = 180-50 = 130
> ...



i sense voo-doo magic afoot Sa967St is hiding something most definitely.


----------



## Robert-Y (Apr 24, 2009)

@qqwref: Maybe I'm wrong but can't you just substitute around to get two equations with x and any other unknown, then use any simultaneous equations solving method to get x?


----------



## keemy (Apr 24, 2009)

@Robert-Y

y+z = 180-50 = 130
x+w = 180-40 = 140
x+z = 180-70 = 110
y+w = 180+180-110-50-40 = 160

goes to like
w+x+y+z=270

if you add the first 2 or 2nd 2 equations which is obvious from where they came from.


----------



## nitrocan (Apr 24, 2009)

Here's my solution:

The bold lines are the equal ones.


----------



## keemy (Apr 24, 2009)

oh cool an actual geometric solution is about as ugly as mine XP


----------



## Sa967St (Apr 24, 2009)

oops, yeah I did leave out something. I had more eqations, then I solved for x using them and the other ones . I guess I accidentally deleted them before I submitted the post @[email protected]


----------

