bonanova Posted September 12, 2008 Report Share Posted September 12, 2008 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.You can add a line or two if you think it will help. Quote Link to comment Share on other sites More sharing options...
0 dms172 Posted September 12, 2008 Report Share Posted September 12, 2008 x=40 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 as far as i can tell, x could be anywhere from 1 degree to 129 degrees... Quote Link to comment Share on other sites More sharing options...
0 bonanova Posted September 12, 2008 Author Report Share Posted September 12, 2008 No correct answers so far. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 No correct answers so far. Isn't the law of sines dirrived from simple germetrical properties? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 This is way harder than it looks, and it's going to bug me all day. I'll get back to you. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 as far as i can tell, x could be anywhere from 1 degree to 129 degrees... Well I've narrowed it down a bit fartherx is less than 70 degrees Still working...maybe I'll have a diagram to post later. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 (edited) 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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 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:is 40 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 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! Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 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. 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). Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 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. 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). but so far, none of the lines I've added tell me anything I don't already know without the lines Quote Link to comment Share on other sites More sharing options...
0 Nikyma Posted September 12, 2008 Report Share Posted September 12, 2008 Yeah, I looked at this for a good while last night. Couldn't get it. Will try again after work. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 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... Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 EDIT: Classification of AnglesDAE=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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 x=50? Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 (edited) x=90? oops took too long to do a look over...yeah i'm pretty sure x =90° Edited September 12, 2008 by Impervious Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 Why do you think x=90??? Can u please explain... Thx Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 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. 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 with confirming my points with the previous posts x= 30 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 (edited) Well, I don't know the right answer yet, but I can prove this one is incorrect. 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 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 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 with confirming my points with the previous posts x= 30 you're right. I just plugged in the numbers. Cherry Lane ignore my old angles Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 Cherry Lane ignore my old angles 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. Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 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 ) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted September 12, 2008 Report Share Posted September 12, 2008 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. Quote Link to comment Share on other sites More sharing options...
0 Prof. Templeton Posted September 12, 2008 Report Share Posted September 12, 2008 (edited) 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 Quote Link to comment Share on other sites More sharing options...
Question
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
Have fun.
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