The following link will take you to a file that has a screenshot of my problem. I'm not allowed to use trig.
https://drive.google.com/file/d/0B9q...=sharing(Rofl)
The following link will take you to a file that has a screenshot of my problem. I'm not allowed to use trig.
https://drive.google.com/file/d/0B9q...=sharing(Rofl)
I'm not sure why you would "label the vertex at the 'x' F" when it is already labeled "E". Oh, I see. DenisB is not referring to the "x degrees" that you are to determine, he is referring to the point where the two line segments, AE and BD, intersect.
I would use the fact that the measures of the angles in a triangle sum to 180 degrees repeatedly. You are given that angle BAE has measure 60 degrees and angle ABD has measure 50, angle AFB has measure 180- 60- 50= 70 degrees. By the 'vertical angle theorem', angle DFE is also 70 degrees. We have that angle ABE has measure 50+ 30= 80 degrees so angle 180- 60- 80= 40 degrees. There are many more triangles- keep doing that.
I did that, and this is where I got. https://docs.google.com/a/tsc.k12.in...it?usp=sharing
Lol I'm kinda bad at geometry cause I started a month and a half ago but heres my steps
1. Find Triangle CEA, we know that two of the angles are 20° each so that leaves us with the last angle being 140° we'll call this angle <E+X
2. Find Triangle CDB, we have two angles which are 20° and 30°, 180-50 is 130° (the angle that is 130 is <CDE and <CDB)
3. Let's call angle EDB, "Y"
4. When we compare Triangle CEA and Triangle CDB, we want to focus on the third sides, so 130° and 140°
CEA is comprised of X + __ = 140
CDB is comprised of Y + __ = 130
5. Then I realized that you can do substitution, because the difference between 140 and 130 is ten, Y= X-10
6. Let us focus on the center triangle (the one with X in it) it's Y + X + 70 = 180
Substitute Y for X-10,
X-10+X+70= 180 -> 2x-10 = 110 -> 2x=120 -> X=60
7. I know I'm late and you probably already discussed this question in your class already :/
AEB = 40° because 60+30++50= 140, 180-140 = 40
ADB = 50° because 60+20+50= 130, 180-130 = 50
DFE = 70° because 60+50 = 110, <F is 70° and by vertical angles it also = 70
What other reasoning's did you use?
The only other way I've seen to get 30 is get triangle BDC, assume <C is 20, 180-50 = 130 and subtract 50 to get 80, now angle D in triangle DEF is 80+70+X = 180