Find 'x'. I am doing geometry and i got stuck in this question.

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I tried and all i got was x+z=130, a+z=140 and a+y=100; where z=angle<EDF and y = angle <DEP and a= angle<DEF; P is the point between B and E where line AD insersects
 
There is one more equation that you could get. Then solve the system of equations. You will be surprised at what you get.
 
I've seen many problems similar to this; the angles alone are not sufficient to solve for the unknown angle. It is clear by construction that angle x is determined by the given data, but something involving the actual intersections (and related lengths) is needed to find it.

I expected, based on such problems, that x might turn out to be some nice number like 10 or 20 degrees that could be found by some special insight; but I just constructed it on GeoGebra and found that it is not anything special. I'd try trigonometry next. There are a couple isosceles triangles here, which may help.
 
Dr.Peterson said:
I've seen many problems similar to this; the angles alone are not sufficient to solve for the unknown angle. It is clear by construction that angle x is determined by the given data, but something involving the actual intersections (and related lengths) is needed to find it.

I expected, based on such problems, that x might turn out to be some nice number like 10 or 20 degrees that could be found by some special insight; but I just constructed it on GeoGebra and found that it is not anything special. I'd try trigonometry next. There are a couple isosceles triangles here, which may help.
Click to expand...
So you are saying that 0<x<90 is not correct?
 
Jomo said:
So you are saying that 0<x<90 is not correct?
Click to expand...
The outer triangle is defined by 3 angles (well, side lengths are arbitrary). Then you add the 2 lines from the top and left vertices. And finally the 3rd line. No ambiguity anywhere.
 
I think @hoosie has the correct answer.

I tried using the simultaneous equation method suggested earlier but, for me at least, it didn't lead to a solution. ie not linearly independent.

I spotted quite an elegant alternative method (after staring at it for a long time). Using the symbols in Hoosie's diagram...
  1. Let length CD=d
  2. BC = d*tan(50)
  3. AC = BC*tan(30) = d*tan(30)*tan(50)
  4. CE = AC*tan(10) = d*tan(10)*tan(30)*tan(50)
  5. angle x = tan-1(CE/CD) = tan-1(tan(10)*tan(30)*tan(50)) ≃ 6.9175
 
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