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Off topic => Break Room => Topic started by: Cutter on January 04, 2019, 12:32:26 PM

Title: Solve
Post by: Cutter on January 04, 2019, 12:32:26 PM
Can you solve the problem without a computer ??

Title: Solve
Post by: Sparky_NY on January 04, 2019, 06:06:05 PM


 


Can you solve the problem without a computer ??

 




No,   only one point is shown,   need to have a reference point to get a measurement.    There are at least two possibilities...     the point labeled zero and the horizontal center line.

Title: Solve
Post by: Cutter on January 04, 2019, 06:42:40 PM


No,   only one point is shown,   need to have a reference point to get a measurement.    There are at least two possibilities...     the point labeled zero and the horizontal center line.




 


This is solvable , no trick involved
Title: Solve
Post by: bruski on January 04, 2019, 09:22:30 PM

Is it .500


Title: Solve
Post by: Sparky_NY on January 04, 2019, 10:34:34 PM


 


This is solvable , no trick involved

 




Not saying its a trick,   its only half a question.        \"Location\" from what?????      There IS more than one choice.         I am guessing you are thinking the point marked \"zero\" but it could just as easy be the perpendicular distance from the centerline.


 


Or.. is it XY coordinates you are looking for?


Title: Solve
Post by: Roller on January 05, 2019, 10:10:02 AM
X .584 Y .7334 ?
Title: Solve
Post by: Cutter on January 05, 2019, 11:00:27 AM


Is it .500




No Sir

Title: Solve
Post by: Cutter on January 05, 2019, 11:00:49 AM


X .584 Y .7334 ?




No Sir

Title: Solve
Post by: Roller on January 05, 2019, 12:45:58 PM
X .522 Y .778 ?
Title: Solve
Post by: maccrazy2 on January 05, 2019, 01:08:38 PM
.500, .8125
Title: Solve
Post by: Cutter on January 05, 2019, 01:13:54 PM


.500, .8125




No Sir

Title: Solve
Post by: Cutter on January 05, 2019, 01:17:39 PM


X .522 Y .778 ?




 


Yes Sir,

How did you solve it ?

Title: Solve
Post by: Roller on January 06, 2019, 09:06:45 AM

Calculated the angle and hypotenuse for the 1.00 x .375 triangle.  With three sides known of the middle triangle, calculated the angle.  Added the two angles and subtracted the result from 90. degrees and used the resulting angle to come up with X and Y.


 


Frank


Title: Solve
Post by: maccrazy2 on January 06, 2019, 09:32:50 PM
I sure wish I could go to school for this stuff. Unfortunately my schedule will not allow it.
Title: Solve
Post by: bruski on January 07, 2019, 12:47:19 PM

Ohhhh that location.


 


bruski


Title: Solve
Post by: Cutter on January 07, 2019, 01:21:15 PM

Solution as Roller explained  


Title: Solve
Post by: Cutter on January 17, 2019, 02:16:28 PM
This one requires less Neurons

Title: Solve
Post by: bruski on January 17, 2019, 07:32:08 PM

34.209 degrees


 


bruski


Title: Solve
Post by: Roller on January 18, 2019, 10:16:42 AM
I got 34.2093 degrees.


Frank
Title: Solve
Post by: Cutter on January 18, 2019, 06:28:12 PM


34.209 degrees


 


bruski




How did you solve it ?

Title: Solve
Post by: bruski on January 19, 2019, 12:33:41 AM

I can\'t draw it but it involved a few triangles from the tangents where the straight lines meet up with the diameters. I make for a terrible math teacher I know it.


 


bruski


Title: Solve
Post by: Cutter on January 19, 2019, 10:53:42 AM
There’s usually more than one approach

for solving these, here’s mine

Title: Solve
Post by: Cutter on January 20, 2019, 11:49:51 AM

Last One


Title: Solve
Post by: bruski on January 20, 2019, 12:55:27 PM

Is it 38.577 degrees? I have been watching Joe Pieczynski channel which he is doing a series on rotary tables and there geometry right now.


 


bruski


Title: Solve
Post by: Cutter on January 21, 2019, 03:29:01 PM
Two triangles 

Title: Solve
Post by: Roller on January 21, 2019, 04:53:50 PM
I got 38.4095 degrees.


Frank
Title: Solve
Post by: maccrazy2 on January 21, 2019, 09:13:19 PM
I have been watching Joe’s videos as well. I like that the videos are to the point and generally cover one single subject.
Title: Solve
Post by: Cutter on January 22, 2019, 02:09:37 PM
My Trig table book, used every day when I

was programming NC code with Quill & Ink

Title: Solve
Post by: Roller on January 22, 2019, 07:00:21 PM

Wow, you have the new book!  Mine is dated 1962, LOL!


 


Frank


Title: Solve
Post by: Cutter on February 06, 2019, 05:52:29 PM

OK , last one again


Title: Solve
Post by: bruski on February 10, 2019, 12:01:04 AM

Is it 20 degrees


 


bruski


Title: Solve
Post by: Cutter on February 10, 2019, 11:26:35 AM


Is it 20 degrees


 


bruski




No Sir,


 


 

First solve angle A  by adding B to C and subtracting from 180°

Title: Solve
Post by: Cutter on February 14, 2019, 07:31:40 PM
Next the triangle with two known angles 50° & 60° leaving 70°

for the third angle

Title: Solve
Post by: bruski on February 16, 2019, 11:18:22 PM

The more that I stare at the triangles the more it looks like a pyramid in 3D.


 


bruski


Title: Solve
Post by: Cutter on February 17, 2019, 11:04:44 AM

 When two lines intersect they make vertical angles


Title: Solve
Post by: Cutter on February 20, 2019, 11:05:39 AM
One more step gives the answer

Title: Solve
Post by: bruski on February 20, 2019, 05:02:20 PM

Anybody!!!!


Title: Solve
Post by: Sparky_NY on February 20, 2019, 06:26:55 PM


Anybody!!!!




Ok    30 degrees

Title: Solve
Post by: Stirlingking on February 20, 2019, 11:30:31 PM


Anybody!!!!




 


29° 55\'.  If that\'s not correct I will give up and wait for 


the answer and solution. My brain can\'t take anymore. 


 


Stirlingking 

Title: Solve
Post by: Cutter on February 21, 2019, 01:47:48 PM


Ok    30 degrees




 


The solution is more important than the answer
Title: Solve
Post by: Cutter on February 23, 2019, 10:59:01 AM

By rotating the purple line 50° you will be subtracting 50°

from 130° leaving 80 ° for the second angle.


 


Title: Solve
Post by: Sparky_NY on February 23, 2019, 02:07:02 PM

What allows you to rotate the purple line 50 deg?     Why not 40, 55, or any random number.      What geometric rule allows you to rotate it 50 degrees?


 


The angle labeled 50 degrees does not bear any relationship to the angle you make by rotating the line 50 deg that I I can see.


 


PS   I did subtract the 50 deg from the 130 when I did it BUT later looking at it again decided that was not valid.


Title: Solve
Post by: Cutter on February 23, 2019, 07:26:16 PM


What allows you to rotate the purple line 50 deg?     Why not 40, 55, or any random number.      What geometric rule allows you to rotate it 50 degrees?


 


The angle labeled 50 degrees does not bear any relationship to the angle you make by rotating the line 50 deg that I I can see.


 


PS   I did subtract the 50 deg from the 130 when I did it BUT later looking at it again decided that was not valid.




 


Do you agree the 50° has already been established as one angle of that triangle ?


If you agree, why think of a random number ?  If not, what\'s your solution.


Title: Solve
Post by: Sparky_NY on February 23, 2019, 09:18:17 PM


Do you agree the 50° has already been established as one angle of that triangle ?


If you agree, why think of a random number ?  If not, what\'s your solution.




3 different triangles all meet at point F.     ONE of them has a established 50 deg angle. (and its not the triangle with \"X\" that we are looking for)       I\'m not sure which triangle your are referring to by \"that\" triangle.


 


I do not follow your explanation at all.


 


As I said, my solution was to subtract the 50deg from the 130 deg,  thats how I came up with 30 degrees for a previous answer based on 180deg total for all angles in the triangle.   However, when looking later, I could not see why I felt it was a valid operation at the earlier time.      I still don\'t see any geometric rule that allows that.


Title: Solve
Post by: Stirlingking on February 23, 2019, 10:47:25 PM


3 different triangles all meet at point F.     ONE of them has a established 50 deg angle. (and its not the triangle with \"X\" that we are looking for)       I\'m not sure which triangle your are referring to by \"that\" triangle.


 


I do not follow your explanation at all.


 


As I said, my solution was to subtract the 50deg from the 130 deg,  thats how I came up with 30 degrees for a previous answer based on 180deg total for all angles in the triangle.   However, when looking later, I could not see why I felt it was a valid operation at the earlier time.      I still don\'t see any geometric rule that allows that.




 


I concur.  Rotating line segment AB 50° Clock Wise to make it fall on side EF (co-linear?) 


would require assuming angle AFE (vertex at F) was 50°, leaving me with no valid solution. 

Given triangle ABE was a 20° 20° 140° Isosceles triangle, ie having 2 equal sides, 

assigned length values to the equal sides and solved for lengths of the legs of the triangles 

to finally arrive at the last step.  Since the little triangles were scalene and not right triangles, 

I had to do the a2 + b+ 2 AB etc for each triangle which is why my brain is fried.  

My answer being 29° 55\' and not 30° is probably due to rounding errors. The slide rule I used 

in basic electronics would have nailed it. 

I am anxiously waiting for more learning opportunities from the Forum.  

 

Stirlingking    

Title: Solve
Post by: Sparky_NY on February 24, 2019, 07:08:36 AM

This morning I notice something.....    along with the purple line,  there is also a purple arrow on the drawing.       Rotating the line pointed to by the purple arrow 50 degrees WOULD be valid because that angle is defined.


 


Although Jerry\'s post said \"rotating the purple line\",     perhaps he meant the purple arrow?        I am guessing the purple arrow was put there for some reason.


Title: Solve
Post by: Cutter on February 24, 2019, 10:25:07 AM


 


I concur.  Rotating line segment AB 50° Clock Wise to make it fall on side EF (co-linear?) 


would require assuming angle AFE (vertex at F) was 50°, leaving me with no valid solution. 

Given triangle ABE was a 20° 20° 140° Isosceles triangle, ie having 2 equal sides, 

assigned length values to the equal sides and solved for lengths of the legs of the triangles 

to finally arrive at the last step.  Since the little triangles were scalene and not right triangles, 

I had to do the a2 + b+ 2 AB etc for each triangle which is why my brain is fried.  

My answer being 29° 55\' and not 30° is probably due to rounding errors. The slide rule I used 

in basic electronics would have nailed it. 

I am anxiously waiting for more learning opportunities from the Forum.  

 

Stirlingking    

 




 


Stirlingking,

This problem doesn’t require any trigonometry formulas.

Simple math with round numbers , finding three numbers equaling 180°

Title: Solve
Post by: Cutter on February 24, 2019, 10:33:04 AM


3 different triangles all meet at point F.     ONE of them has a established 50 deg angle. (and its not the triangle with \"X\" that we are looking for)       I\'m not sure which triangle your are referring to by \"that\" triangle.


 


I do not follow your explanation at all.


 


As I said, my solution was to subtract the 50deg from the 130 deg,  thats how I came up with 30 degrees for a previous answer based on 180deg total for all angles in the triangle.   However, when looking later, I could not see why I felt it was a valid operation at the earlier time.      I still don\'t see any geometric rule that allows that.




 


The corner of \"that\" triangle with 50 in the circle


The geometric rule is a straight line equals 180°

Title: Solve
Post by: Sparky_NY on February 24, 2019, 12:27:08 PM

Still many unanswered questions.....    purple line, purple arrow,  exactly which line is proposed to be rotated...  etc etc


 


I am going back into watch mode.


Title: Solve
Post by: Cutter on February 24, 2019, 01:38:56 PM
Maybe this will bring a Bingo moment .

Title: Solve
Post by: bruski on February 24, 2019, 05:40:22 PM

What language are you guys speaking?


 


bruski


Title: Solve
Post by: Stirlingking on February 24, 2019, 11:10:35 PM
I stared at the triangle for hours, drew triangles and quadrilaterals (4 angles add up to 360°) 

for more hours, always getting stuck at the angles at \"F\". 

I couldn\'t come up with any theorem, axiom, postulates or corollaries that would give 

angle AFE the value of 50°, the same as angle BFC which would have made side BFA 

colinear with FE when rotated 50° clockwise, allowing \"x\" to be solved.   

Without a theorem for authority, rotating side BFA 50° clockwise and be colinear with 

FE would be \"over determining\", in which one condition can be met, but not both, in this 

case unless angle AFE was determined to be 50° beforehand.  

(If it turned out angle AFE was actually 49° or 51°, forcing the two lines to be the same, 

would be \"over determining\" and the answer incorrect, but in this case it wasn\'t.

Unable to find the angles at point \"F\" by the method given, I did a workaround to find the 

elusive angles.  If nothing else, I verified \"X\" to be 30°.  

After having said my piece, and then some, I might lay low and get something done in my shop. 

Stirlingking 

Title: Solve
Post by: Cutter on February 25, 2019, 11:07:29 AM


What language are you guys speaking?


 


bruski




 


Junior high math,1953

 


I can’t speak for Stirlingking ☺☺

Title: Solve
Post by: Sparky_NY on February 25, 2019, 12:31:56 PM


 


Junior high math,1953

 


I can’t speak for Stirlingking ☺☺


 




Stirlingking\'s responses are the sort of detail we were required to give in order to pass Geometry.     We had to cite the specific axiom, theorem etc. that validated each operation as legal.  


 


I completely concur with Stirlingking\'s post.       I think a communications gap is what is leading to much of the confusion here, some of it has been rather cryptic.


 


I took and passed Geometry first in 1966 and later in college , its still very fresh in my memory.  


Title: Solve
Post by: bruski on February 25, 2019, 04:54:03 PM

Isn\'t Geometry the study of rocks? I\'m kidding, I meant rock formations.


 


bruski


Title: Solve
Post by: Cutter on February 25, 2019, 07:00:04 PM
You’ve got got this Mr.B

Title: Solve
Post by: Stirlingking on February 25, 2019, 08:54:33 PM


 


Junior high math,1953

 


I can’t speak for Stirlingking ☺☺


 




 

Here I go submitting another long rambling post. 

I took geometry in the 10th and 11th grade, the same class twice, probably getting a C. 


Out of Voc Tech school I worked at several hole-in-the-wall shops in which it was possible 

the boss couldn\'t make payroll on payday.  I made it a point to be proficient at shop math, 

and put out good work so it wasn\'t my doing for money not coming in. Then I went to Bo**ng 

where I took all the off hour classes could , but eschew have little use for academia.  

No offence to those out there with degree or two. 

OK, I\'m a control freak. I can\'t help it, I was born that way.

             ...(cymbal crash)... 

                  ...(silence)...  

Darn, no one ever gets it! 

Stirlingking 

Title: Solve
Post by: bruski on February 26, 2019, 12:34:34 AM


 


You’ve got got this Mr.B

 




Is that a petrified math equation from the Mesozoic era time period?


 


bruski

Title: Solve
Post by: Cutter on February 26, 2019, 11:15:57 AM
I’ve hit rock bottom on this thread, so

I’ll try for more progress with the GG

Title: Solve
Post by: 4171 on February 26, 2019, 12:39:49 PM
Lol Cutter. I know I’m confused 😐

Mike
Title: Solve
Post by: Stirlingking on February 27, 2019, 04:30:55 PM


 


I’ve hit rock bottom on this thread, so

I’ll try for more progress with the GG

 




 


Maybe that rock is from the Jurassic period.  WMBOWS 


Wash My Brain Out With Soap. 


I would submit a witty post, but my fingers are talked out. 


Stirlingking 


 


\"This year will go down in history. For the first time, a civilized nation has full gun registration! 

Our streets will be safer, our police more efficient, and the world will follow our lead into the future!\" 

Adolph Hitler.....1934