# I am unable to solve the angle x in the diagram.  Please solve the value x for me.

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embizze | Certified Educator

There is not a singular answer for the value of x.

In the diagram as given, mark the intersection of `bar(BD),bar(EA) ` as F.

Then:

`m/_AFB=m/_ EFD=50 `

`m/_DFA=m/_EFB=130 `

`m/_BDA=40,m/_AEB=30 `

All of these are results of looking at each triangle and using the fact that the angle sum of a triangle is 180 degrees.

Now let `x=/_DEF,y=/_EDF,z=/_DEC,w=/_CDE `

Again applying the angle sum of triangles we get the following:

x+y=130
x+z=150
y+w=140
z+w=160

This system is consistent and dependent -- there are multiple solutions. The solutions can be described as follows:

For 0<x<130
y=130-x
z=150-x
w=x+10

For example, let x=10; then y=120,z=140, and w=20. These angles all work in the diagram.

Or we could let x=20; then y=110,z=130,w=30. These angles also work in the diagram.

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There are multiple solutions for this problem: if 0<x<130, then the angles as described above will be y=130-x, z=150-x, and w=x+10.

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The diagram indicates that it is drawn to scale, indicating a small value for x. But the given information does not require a small value for x: x can take on any value between 0 and 130 degrees (not including 0 or 130.)