Solve

[Cutter]

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

from 130° leaving 80 ° for the second angle.

 

[Sparky_NY]

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.

[Cutter]

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.

[Sparky_NY]

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.

[Stirlingking]

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, 

I 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 + b² + 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    

[Sparky_NY]

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.

[Cutter]

 

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, 

I 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 + b² + 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°

[Cutter]

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°

[Sparky_NY]

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.

[Cutter]

Maybe this will bring a Bingo moment .

[bruski]

What language are you guys speaking?

 

bruski

[Stirlingking]

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 

[Cutter]

What language are you guys speaking?

 

bruski

 

Junior high math,1953

 

I can’t speak for Stirlingking ☺☺

[Sparky_NY]

 

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.  

[bruski]

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

 

bruski

[Cutter]

You’ve got got this Mr.B

[Stirlingking]

 

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 I could , but I eschew have little use for academia.  

No offence to those out there with a 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 

[bruski]

 

You’ve got got this Mr.B

 

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

 

bruski

[Cutter]

I’ve hit rock bottom on this thread, so

I’ll try for more progress with the GG

[4171]

Lol Cutter. I know I’m confused 😐

Mike