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

Asked on by fredha2013

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embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

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.)

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embizze | High School Teacher | (Level 2) Educator Emeritus

Posted on

Using a geometry program, the value for x would be 20. The other values work for the angles, but will not work for the side lengths. (Using other values, the segments do not intersect as in the drawing.)

Triangle DEC is similar to triangle EFB where F is the intersection of BD and AE. Then angle CED is 130; since angle BEA is 30 then x=20.

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