Thursday, August 14, 2008

A Little Trigonometry



Find the angle , one who get it fastest wins :)

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17 Comments:

Anonymous BigMo said...

15

August 14, 2008 10:27 PM  
Anonymous Anonymous said...

Not solvable.
Without knowing the relevant points of D or E, the center triangle can be any type of triangle.


~Cybersurf~

August 15, 2008 1:23 AM  
Anonymous Anonymous said...

forget what I said!

~Cybersurf~

August 15, 2008 1:25 AM  
Blogger mfbabin said...

B=20
D=110
E=50
Is it right?

August 15, 2008 7:13 AM  
Anonymous Anonymous said...

CBD = 80 60 + (20)
BCA = 80 50 + (30)
CDE = 60
DEB = 50

100% sure it is correct. I have all angles filled in.

August 15, 2008 9:13 AM  
Anonymous Anonymous said...

The anonymous right above me is correct. thats what i got

August 15, 2008 9:26 AM  
Anonymous Anonymous said...

the congruent angle markers need to be removed if your going to call the inner part of b and c different things, as well as angle a

August 15, 2008 2:54 PM  
Anonymous Anonymous said...

the guy 3 above me, explain yourself. you cant just say that and prove it.

August 16, 2008 7:45 AM  
Anonymous Travis said...

Angle DEB can be 50 degrees.
It would make triangle ADE a 20 70 90 triangle, and triangle CDE a 30 60 90.
But Angle DEB can also be 30.
This would make ADE a 10 50 110 triangle and CDE a 30 80 70.
If fact if you print the graphic, it's pretty clear that CED is not a right angle, so DEB as 50 doesn't really work visually even if it works mathmatically.

August 16, 2008 11:26 AM  
Anonymous Anonymous said...

guy above me, the drawing probly aint to scale, and you ADE is missing 10 degrees

August 16, 2008 11:59 PM  
Anonymous Travis said...

Right, ADE is 20 50 110. I typed it wrong, sorry.
I agree we can't assume the drawing isn't to scale. This reinforces the point that the very first poster was making and that I was making: There can be multiple solutions based on the information given.

August 17, 2008 12:57 PM  
Anonymous Anonymous said...

not all the angles could have been filled in. You can't just assume or hold your protractor up to the computer screen to solve this.

August 23, 2008 2:02 AM  
Anonymous Anonymous said...

right there

August 23, 2008 1:40 PM  
Anonymous Anonymous said...

Based on the information given, there is only one right answer. The drawing is not to scale. It helps to define a point F at the intersection of CD and BE. The angles are as follows:

ABC = 80°
ABE = 20°
ACB = 80°
ACD = 30°
ADC = 130°
ADE = 50°
AEB = 140°
AED = 110°
BAC = 20°
BCD = 50°
BDC = 50°
BDE = 130°
BEC = 40°
BED = 30°
BFC = 70°
BFD = 110°
CBE = 60°
CDE = 80°
CED = 70°
CFE = 110°
DFE = 70°

The correct answer is 30°

August 26, 2008 12:39 PM  
Anonymous Anonymous said...

how can the 2 lines be parrallel (spelt wrong :D) if its a triangle ??? hmmm.... and answer 30* ??

August 31, 2008 10:31 AM  
Anonymous Anonymous said...

Angle DEB is definitely 30 degrees. All the earlier posters seem to have gotten it wrong...the last few, though are correct.

Good puzzle! Does anyone have any more like this to post?

Rob Miller

December 29, 2008 10:14 AM  
Anonymous Anonymous said...

agreed
DEB = 30
EDF = 80
DFE = 70

Not sure how you guys done it but I had to subsitute a value for BC of 100m then went from there.

November 3, 2009 8:43 PM  

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