Using only elementary geometry, determine angle x. Provide a step-by-step proof.
You may use only elementary geometry, such as the fact that the angles of a triangle add up to 180 degrees and the basic congruent triangle rules (side-angle-side, etc.). You may not use trigonomery, such as sines and cosines, the law of sines, the law of cosines, etc. There is a review of elementary geometry below.
This is the hardest problem I have ever seen that is, in a sense, easy. It really can be done using only elementary geometry. This is not a trick question.

- anonymous

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- katieb

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- anonymous

wow, it is hard... i don't even know where to begin...

- anonymous

can any one answer my question

- anonymous

it is very simple use elementary geometry method

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## More answers

- anonymous

@dpaInc give a medal man

- anonymous

Is that the only information?

- lgbasallote

no drawings?

- anonymous

yah

- anonymous

Its like asking to find the area of a geometrical object without being told what object is that

- lgbasallote

ill try and solve this...then call on the geniuses lol =))

- anonymous

here's a support

##### 1 Attachment

- lgbasallote

the two base angles total 160 so the last angle is 20...that's a starter

- anonymous

give medal@lgbasallote

- experimentX

quite easy

- anonymous

give solution

- NotTim

hey, why not solve for all the angles? Thats what I do

- experimentX

okay 3 min

- lgbasallote

pretty sure transversals arent allowed as well?

- anonymous

Here is everything you need to know to solve the above problems.
Lines and Angles: When two lines intersect, opposite angles are equal and the sum of adjacent angles is 180 degrees. When two parallel lines are intersected by a third line, the corresponding angles of the two intersections are equal.
Triangles: The sum of the interior angles of a triangle is 180 degrees. An isosceles triangle has two equal sides and the two angles opposite those sides are equal. An equilateral triangle has all sides equal and all angles equal. A right triangle has one angle equal to 90 degrees. Two triangles are called similar if they have the same angles (same shape). Two triangles are called congruent if they have the same angles and the same sides (same shape and size).
Side-Angle-Side (SAS): Two triangles are congruent if a pair of corresponding sides and the included angle are equal.
Side-Side-Side (SSS): Two triangles are congruent if their corresponding sides are equal.
Angle-Side-Angle (ASA): Two triangles are congruent if a pair of corresponding angles and the included side are equal.
Angle-Angle (AA): Two triangles are similar if a pair of corresponding angles are equal.

- NotTim

|dw:1333113296540:dw| that allowed?

- NotTim

Ba-FOOP. Rohan lets go another fact-filled comment.

- lgbasallote

ohhh transversal fine...then this is solveable...but difficult to explain withoout drawing

- lgbasallote

the angles of the base triangle is 70-60-50..another starter

- anonymous

@lgbasallote give medal

- lgbasallote

the angle on above 50 would be equal to 50 too according to the vertical theorem or something..so the angles of the triangle above the base triangle is 50-x-and another unknown

- lgbasallote

already did roh

- anonymous

sorry

- lgbasallote

ok...now im stuck...can i call the geniuses now? haha lol

- anonymous

i can give the solutions but i will not do so !! find ans....!!!

- lgbasallote

i know! ill solve the angles around the 50s next...100 + 2a = 360
2a = 260
a = 130...the angles around 50s are both 130..good start

- anonymous

hey all give medals

- anonymous

If memory serves I know at-least 8 ways of solving this one ;)

- anonymous

lets do a a competition regarding problem of the day

- lgbasallote

having known that...the triangle on theleftmost has angles 10-130-40..the rightmost has 20-130-30....ffm knows all o.O hahaha

- anonymous

Well this the much easier version of the problem what I am talking about.

- anonymous

so give the solutions

- anonymous

http://www.arbelos.co.uk/Papers/Triangle-problem.pdf

- lgbasallote

im stuck with the last two angles...the last angle in the topmost triangle...and the supplement of that angle and the one beside it...if that is only known then x is solveable

- anonymous

- anonymous

http://agutie.homestead.com/files/LangleyProblem.html

- anonymous

Proofs may be written informally using plain English. Just be sure to include all the steps in your reasoning, or at least all the key steps. Providing a diagram is very helpful but not required. You can draw a diagram on the computer or you can draw it on paper and then scan it or photograph it with a digital camera.
Please number your steps. This makes it easy for both writer and reader to talk about the steps.
Name each point you use with a letter (ex., say "point A" or simply "A"). Identify lines with two letters (ex., say "line AB" or simply "AB"). Identify triangles with three letters (ex., say "triangle ABC" or "tri ABC" or simply "ABC"). Identify angles with three letters, vertex in the middle (ex., say "angle ABC" or "ang ABC" or simply "ABC").
If you don't provide a diagram, you will need to describe the named points with words (ex., say "the intersection of AE and DB is G "). Even if you provide a diagram, you must define with words each new line that you draw, in order (ex., say "Draw a line through C perpendicular to AB intersecting AB at H").
Justify all key steps (ex., say "AC=BC because ABC is isosceles"). You may omit the justification when simply calculating an angle based on any of these simple rules: triangle angles sum to 180, supplementary angles sum to 180, opposite angles are equal.

- anonymous

You may have seen this:
1/81 = 0.012345679012345679...
But you may not have seen this:
1/243 = 0.004115226337448559...
How does the pattern continue? What is special about the number 243? (Hint: find its factors.) What causes the pattern? Are there analogous numbers in other bases?
Source: I found this in Surely, You're Joking Mr. Feynmann by Richard Feynmann.
Imaginary Powers
The concept of imaginary powers is very strange, even if you are comfortable with imaginary numbers (ex., √-1), negative powers (ex., x-1 = 1/x), and fractional powers (ex., x1/2 = √x).
You may have seen this famous, beautiful, and strange equation that relates the most important transcendental numbers, π (3.14159...) and e (2.71828...), with i = √-1, the imaginary square root of -1:
eiπ = -1
This is a special case (with x = π) of Euler's formula:
eix = cos(x) + i sin(x)
Here's another not-as-famous strange equation involving an imaginary power:
ii = e-π/2 = 0.207879576...
This is a special case (with n = 0) of this formula:
ii = e(-π/2 + 2πn)
According to this formula, ii has an infinite number of values! For example, another value (with n = 1) is:
ii = e3π/2= 111.317778...

- anonymous

give this ques ans to get medals

- anonymous

does it have an infinite number of solutions?

- anonymous

@anonymoustwo44 cant say

- anonymous

@saifoo.khan give a solution to this

- anonymous

@Mani_Jha give a solution to this

- anonymous

@experimentX give ans......

- experimentX

i would like to take my statement back as "easy" ... but i am not backing off from the question ...

- anonymous

but is it possible? in this case

- lgbasallote

@experiment use the data i have collected :) it's already near

- anonymous

So are we still on that Geometry problem?

- experimentX

answer might be 50

- anonymous

\[\frac{1}{243}= \frac{1}{3}\times\frac{1}{3}\times\frac 1 3\times\frac{1}{3}\times\frac{1}{3} = .3\bar3 \times .3\bar3 \times .3\bar3 \times .3\bar3 \times .3\bar 3\]

- experimentX

??? how did you get that?

- experimentX

@Rohangrr are you there?

- anonymous

Is the 'you' me?or Rohan?

- anonymous

It's probably Rohan

- experimentX

well, if you can explain ...

- anonymous

come on guys give answer to y question!!

- anonymous

What is the question first of all? Why did you post some random pellet within the thread? Are we still on that Geometry problem?

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