[bonanova]

Triangle ABC is isosceles.

Find angle x.

Do not assume the figure is drawn to scale.

Do not use the law of sines.

Use simple geometric reasoning, like

• 
  
• Isosceles triangles have two equal angles and sides.
  
• The angles of a triangle sum to 180 degrees.
  
• Things like that.
  

Have fun.

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You can add a line or two if you think it will help.

[dms172]

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x=40

[Guest]

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as far as i can tell, x could be anywhere from 1 degree to 129 degrees...

[bonanova]

No correct answers so far.

[Guest]

  bonanova said:

No correct answers so far.

Isn't the law of sines dirrived from simple germetrical properties?

[Guest]

This is way harder than it looks, and it's going to bug me all day. I'll get back to you.

[Guest]

  Marinja said:

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as far as i can tell, x could be anywhere from 1 degree to 129 degrees...

Well I've narrowed it down a bit farther

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x is less than 70 degrees

Still working...maybe I'll have a diagram to post later.

[Guest]

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DAE=10

DBE=20

A_D=130

B_E=130

A_B=50

D_E=50

ADB=40

CDB=140

AEC=150

BEA=30

Updates coming once I do more figuring (unless someone beats me to it)

EDIT: Classification of Angles

Edited September 12, 2008 by Mumbles140

[Guest]

  Mumbles140 said:

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DAE=10

DBE=20

A_D=130

B_E=130

A_B=50

D_E=50

ADB=40

CDB=140

AEC=150

BDA=30

Updates coming once I do more figuring (unless someone beats me to it)

I concur, with one exception:

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is 40

[Guest]

Yah, that's what I corrected...BDA is the same angle as ADB, which I had already listed. I made the correction, but thanks though for keeping me sharp!

[Guest]

There's a hint in the OP hidden in the spoiler tag. I want to add some discussion to that, but I'll hide it as well. I'm guessing the problem requires using the hint to solve.

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There are a few lines I've considered adding. One is to add another point F on leg AC so that AFEB is a trapezoid. This hasn't helped me much. I've also tried duplicating the interior lines (BD, DE, and AE) in mirror image, but that hasn't helped much either. I think you might need to add some elevations, maybe in triangle DCE. I'm working on other options (but also real work that I get paid to do, so expect delays).

[Guest]

  Chuck Rampart said:

There's a hint in the OP hidden in the spoiler tag. I want to add some discussion to that, but I'll hide it as well. I'm guessing the problem requires using the hint to solve.

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There are a few lines I've considered adding. One is to add another point F on leg AC so that AFEB is a trapezoid. This hasn't helped me much. I've also tried duplicating the interior lines (BD, DE, and AE) in mirror image, but that hasn't helped much either. I think you might need to add some elevations, maybe in triangle DCE. I'm working on other options (but also real work that I get paid to do, so expect delays).

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but so far, none of the lines I've added tell me anything I don't already know without the lines

[Nikyma]

Yeah, I looked at this for a good while last night. Couldn't get it. Will try again after work.

[Guest]

I've tried a lot of stuff, but feel like there's something missing.

Isn't there another condition that we cannot assume, but must be in the OP???

Just a guess...

[Guest]

  Mumbles140 said:

EDIT: Classification of Angles

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DAE=10

DBE=20

A_D=130

B_E=130

A_B=50

D_E=50

ADB=40

CDB=140

AEC=150

BEA=30

Updates coming once I do more figuring (unless someone beats me to it)

that's all i've got so far too

[Guest]

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x=50?

[Guest]

  Impervious said:

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x=90?

oops took too long to do a look over...yeah i'm pretty sure x =90°

Edited September 12, 2008 by Impervious

[Guest]

Why do you think x=90???

Can u please explain... Thx

[Guest]

  Quote

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x=90?

oops took too long to do a look over...yeah i'm pretty sure x =90°

Well, I don't know the right answer yet, but I can prove this one is incorrect.

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Draw a line parallel to AB that passes through E. The point where the new line crosses AC is labeled as F. We know that point F is closer than D is to point C along line AC because angl EAB is greater than angle DBA.

Because new line EF is parallel to line AB, we know that angle FEA = EAB = 70. And x is a portion of angle FEA, so x is less than 70.

[Guest]

with confirming my points with the previous posts

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x= 30

[Guest]

  Cherry Lane said:

Well, I don't know the right answer yet, but I can prove this one is incorrect.

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Draw a line parallel to AB that passes through E. The point where the new line crosses AC is labeled as F. We know that point F is closer than D is to point C along line AC because angl EAB is greater than angle DBA.

Because new line EF is parallel to line AB, we know that angle FEA = EAB = 70. And x is a portion of angle FEA, so x is less than 70.

interesting because the angles i came up with were this

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Angles:

ABC = 80°

ABD = 60°

ACB = 20°

ADE = 80°

ADB = 40°

AEB = 30°

AEC = 150°

AED = 90° also angle x

A_B = 50°

A_D = 130°

BAC = 80°

BAD = 10°

BAE = 70°

BDC = 140°

BDE= 40°

BED = 120°

B_E = 130°

CDE = 100°

CED = 60°

DBE = 20°

D_E = 50°

if you could...point out which are incorrect please

Edited September 12, 2008 by Impervious

[Guest]

  solemnraven said:

with confirming my points with the previous posts

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x= 30

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you're right. I just plugged in the numbers.

Cherry Lane ignore my old angles

[Guest]

  Impervious said:

Cherry Lane ignore my old angles

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you're right. I just plugged in the numbers.

And I was just in the process of responding to that, too.

I haven't gotten far enough to comment on the current proposed solution. It passes the reasonableness test, but I'm missing the way to the solution.

[Guest]

  Cherry Lane said:

And I was just in the process of responding to that, too.

I haven't gotten far enough to comment on the current proposed solution. It passes the reasonableness test, but I'm missing the way to the solution.

I actually move away from triangles to calculating trapezoid ADEB

while using the information I knew from

ADB

ABC

BAC

BED

while also knowing what the angle D_E is so it gave me the number that was needed in order to make ∆D_E

Then i began to plug in the obvious number that could solve the puzzle (well incorrectly the first time, i got lost in all my markings)

So yeah, there ya go.

(angles were left out for the sake of those wanting to solve it themselves )

[Guest]

  Impervious said:

I actually move away from triangles to calculating trapezoid ADEB

while using the information I knew from

ADB

ABC

BAC

BED

while also knowing what the angle D_E is so it gave me the number that was needed in order to make ∆D_E

Then i began to plug in the obvious number that could solve the puzzle (well incorrectly the first time, i got lost in all my markings)

So yeah, there ya go.

(angles were left out for the sake of those wanting to solve it themselves )

ADEB is not a trapezoid.

[Prof. Templeton]

  Impervious said:

I actually move away from triangles to calculating trapezoid ADEB

while using the information I knew from

ADB

ABC

BAC

BED

while also knowing what the angle D_E is so it gave me the number that was needed in order to make ∆D_E

Then i began to plug in the obvious number that could solve the puzzle (well incorrectly the first time, i got lost in all my markings)

So yeah, there ya go.

(angles were left out for the sake of those wanting to solve it themselves )

What is angle BED and how did you get you come by the number?

Edited September 12, 2008 by Prof. Templeton