Is the Problem Solvable Without Trigonometry?

[davedave]

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.

 

[ImAnEngineer]

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).

 

[Georgepowell]

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.

 

[uart]

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).

 

[TVP45]

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 180^o.

 

[uart]

Ok I just plugged the numbers and it seems they are chosen so as to make the problem easy.

$$AC = DC \sin(80) / \sin(30) = 2 DC \sin(80)$$

Therefore,

$$BC = DC \sin(80) / \cos(10) = DC$$

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.

 

[ImAnEngineer]

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...)

 

[uart]

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. :tongue2: (j/k)

Sorry I couldn't resist the jab. ;)

 

[ImAnEngineer]

Don't worry, that's close enough for an Engineer. :tongue2: (j/k)

Sorry I couldn't resist the jab. ;)

Haha
It's just a nickname, IamNotReallyAnEngineer

 

[TVP45]

A good engineer always uses tolerances. I'd just say 50^o +/- 15^o. With a large enough hammer, I can make any of those the right answer.

 

[davedave]

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?

 

[ImAnEngineer]

A good engineer always uses tolerances. I'd just say 50^o +/- 15^o. 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

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.

 

[TVP45]

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 180^o. Is that trig?