Is the Problem Solvable Without Trigonometry?

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Discussion Overview

The discussion revolves around a geometric problem involving a quadrilateral with specified angles and the challenge of determining angle BDC. Participants explore whether the problem can be solved without using trigonometry, while also considering various approaches and interpretations of the problem's conditions.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that the problem is solvable, noting that the angles provided allow for the use of properties of triangles, particularly isosceles triangles.
  • Others express skepticism about the solvability without trigonometry, indicating that the problem may be inherently complex.
  • A participant proposes using the sine rule and basic trigonometry to find lengths and coordinates, leading to an estimated angle of BDC around 50 degrees.
  • Another participant calculates angle BDC as approximately 49 degrees, suggesting a possible rounding error in previous estimates.
  • Some participants emphasize the challenge of solving the problem without trigonometry, questioning the boundary between geometry and trigonometric methods.
  • There are humorous exchanges regarding the precision of engineering estimates, with references to tolerances in practical applications.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether the problem can be solved without trigonometry. While some believe it is solvable using geometric principles, others remain uncertain and highlight the difficulty of the task.

Contextual Notes

Some participants note that the problem's conditions may be misinterpreted or inadequately specified, which could affect the approach to solving it. The discussion also reflects varying interpretations of the term "collinear" in the context of the quadrilateral.

davedave
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This is NOT a homework problem.

I went to the library and found this problem in a math book, called "The Solvable and the Unsolvable".

Here is the problem.

Consider a quadrilateral. Its vertices are A,B,C,D.

A is at the top left and B is at the top right lower than A.

C and D are collinear. C is at the bottom right and D is at the bottom left. D is further than A to the left and C is further than B to the right.

Please see the diagram below and IGNORE the dots which are needed to show the locations of the vertices.

.....A
.........B

...D_________________________C

Angle ABC is 160 degrees.
Angle CAB is 10 degrees.
Angle ACB is 10 degrees.
Angle CAD is 30 degrees.
Angle ACD is 70 degrees.
Angle ADC is 80 degrees.

Find angle BDC.

Is it possible to do? If it is, how do you do it? Thanks.
 
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I don't think it is but I might be overlooking something. One thing I would like to point out, though, is that if you isolate triangle ABC you get an isosceles triangle. Might be a clue to the answer to the problem (if there is one).
 
I thought I found the answer! but then I simplified my equation and everything canceled out and I got 0 = 0 :(

Can you give me a link to the book please? I can't find it on Amazon.

If there is no link then the author would be good enough.
 
Yes it is definitely solvable. Start by noting that you are only given angles, but that any similar figure will have the same angles, hence you can choose whatever scale (and origin) that you like.

For example take D as the origin and length CD as unity. First solve for length AC using the sine rule. Then solve for length BC using basic trigonometry and properties of an isosceles triangle. At this point you can find explicit coordinates of point B so it's got to be easy from there. I didn't bother working it out but from my sketch angle BDC looks to be about 50 degrees.

BTW. Are you sure the book stated "C and D are collinear"? That's a silly and unnecessary statement as any two points are always collinear. To tell you the truth I was initially a bit leery of even looking at this problem having read that statement (in case it was misquoted or otherwise inaccurately specified).
 
Last edited:
Definitely can be solved. You even have two extra angle values. When you say C and D are colinear, I interpreted that to mean the line CD is horizontal; else, the figure is not completely solved, although you can get the angle BDC without that.

It's all triangles with 180o.
 
Ok I just plugged the numbers and it seems they are chosen so as to make the problem easy.

[tex]AC = DC \sin(80) / \sin(30) = 2 DC \sin(80)[/tex]

Therefore,

[tex]BC = DC \sin(80) / \cos(10) = DC[/tex]

So it turns out that BCD is also an isosceles triangle. Obviously from here it's dead easy to show that angle BDC=50 degrees.
 
uart said:
For example take D as the origin and length CD as unity.
This helped me to figure it out. I got 49 deg! (might have a rounding off error.. since you have 50...)
 
ImAnEngineer said:
This helped me to figure it out. I got 49 deg! (might have a rounding off error.. since you have 50...)

Don't worry, that's close enough for an Engineer. :-p (j/k)

Sorry I couldn't resist the jab. ;)
 
uart said:
Don't worry, that's close enough for an Engineer. :-p (j/k)

Sorry I couldn't resist the jab. ;)
Haha :wink:
It's just a nickname, IamNotReallyAnEngineer o:)
 
  • #10
A good engineer always uses tolerances. I'd just say 50o +/- 15o. With a large enough hammer, I can make any of those the right answer.
 
  • #11
davedave said:
This is NOT a homework problem.

I went to the library and found this problem in a math book, called "The Solvable and the Unsolvable".

Here is the problem.

Consider a quadrilateral. Its vertices are A,B,C,D.

A is at the top left and B is at the top right lower than A.

C and D are collinear. C is at the bottom right and D is at the bottom left. D is further than A to the left and C is further than B to the right.

Please see the diagram below and IGNORE the dots which are needed to show the locations of the vertices.

.....A
.........B

...D_________________________C

Angle ABC is 160 degrees.
Angle CAB is 10 degrees.
Angle ACB is 10 degrees.
Angle CAD is 30 degrees.
Angle ACD is 70 degrees.
Angle ADC is 80 degrees.

Find angle BDC.

Is it possible to do? If it is, how do you do it? Thanks.

Sorry. There is something I missed when I copied the problem from the library book.

I think what makes it so tough is that in the problem we are supposed to solve it WITHOUT USING TRIGONOMETRY if possible.

Could someone please try to solve it without trigonometry?
 
  • #12
TVP45 said:
A good engineer always uses tolerances. I'd just say 50o +/- 15o. With a large enough hammer, I can make any of those the right answer.
Who cares in the end of the day whether a skyscraper has a 15deg angle in one way or the other :biggrin:

davedave said:
Sorry. There is something I missed when I copied the problem from the library book.

I think what makes it so tough is that in the problem we are supposed to solve it WITHOUT USING TRIGONOMETRY if possible.

Could someone please try to solve it without trigonometry?
You could draw it and measure the angle...
But without trig that's going to be a matter of trial and error I guess.
 
  • #13
davedave said:
Sorry. There is something I missed when I copied the problem from the library book.

I think what makes it so tough is that in the problem we are supposed to solve it WITHOUT USING TRIGONOMETRY if possible.

Could someone please try to solve it without trigonometry?

You'd have to get a mathematician to tell you the boundary between geometry and trig. The only things I used were that three sides make a triangle and the sum of the included angles is 180o. Is that trig?
 

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