This Trig Question had my math teacher stumped!

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yohanson77

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Jan 8, 2007
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Ok I'll get straight into it:

A quadrilateral plot on the ground ABCD is surveyed for redevelopment. It has boundary dimensions: AB = 60.0m, BC = 81,486m, CD =145.0m and has internal diagonal measurements of CA = 124.0m. and BD = 130.0m. Determine the length AD?

All help would be appreciated:)
 
yohanson77 said:
A quadrilateral plot on the ground ABCD is surveyed for redevelopment. It has boundary dimensions:
AB = 60.0m, BC = 81,486m, CD =145.0m and has internal diagonal measurements of CA = 124.0m. and BD = 130.0m. Determine the length AD?
Click to expand...
Those measurements make the problem impossible
Please review and correct!
 
yohanson77 said:
Ok I'll get straight into it:

A quadrilateral plot on the ground ABCD is surveyed for redevelopment. It has boundary dimensions: AB = 60.0m, BC = 81,486m, CD =145.0m and has internal diagonal measurements of CA = 124.0m. and BD = 130.0m. Determine the length AD?

All help would be appreciated:)
Click to expand...

Please double-check that second number. It can't possibly be as indicated if the diagonals are right!
 
It's from the UK, Folks, but I really don't understand why there is a decimal at the ".0" and not at the ",486". Inconsistent notation is harder to work with.

One way is to beat the tar out of it.

Get a coordinate axis.

Put A at the origin.

B is 60 m from A, so it is somewhere on x^2 + y^2 = 60^2.
Pick the solution (0,60).

C is 124 m from A, so it is somewhere on x^2 + y^2 = 124^2.
C is 80.146 m from B, so it is somewhere on x^2 + (y-60)^2 = 81.486^2.
Pick the solution in the first quadrant (69.341,102.800)

D is 130 m from B, so it is somewhere on x^2 + (y-60)^2 = 130^2.
D is 145 m from C, so it is somewhere on (x-69.341)^2 + (y-102.800)^2 = 145^2.
Pick the solution that makes the most sense (80.931,-41.736)

The distance from A to D is now a simple distance formula problem.

There are many solutions. I just directed you to one.

This is messy, but I don't really see why it's tricky. Why were you and your math teacher stuck?
 
Maths Teacher

TKhunny,

There is a drawing to go with it and it is drawn really bad, I think it was the drawing that was throwing us all off.

Thank you for your help, much appreciated:)
 
A quadrilateral plot on the ground ABCD is surveyed for redevelopment. It has boundary dimensions: AB = 60.0m, BC = 81,486m, CD =145.0m and has internal diagonal measurements of CA = 124.0m. and BD = 130.0m. Determine the length AD?

I must assume that the 81,486m number should be either 81.486 or 81.0 (in line with your other numbers.)

If so, you might determine your answer by using Ptolomey's
theorem which states that the sum of the products of the two pairs of opposite sides is equal to the product of the two internal diagonals.

Therefore, 60(145) + 81.486(AD) = 124(130). Solve for AD.

BC cannot be equal to 81,486.
 
TchrWill said:
If so, you might determine your answer by using Ptolomey's
theorem which states that the sum of the products of the two pairs of opposite sides is equal to the product of the two internal diagonals.
Click to expand...
That is a good call. But remember that Ptolemy’s Theorem requires a cyclic quadrilateral, that is, one that can be circumscribed by a circle.

BTW: In South Africa they do use a comma the separate the decimal part. But in the UK a period is used (I studied at the U of Manchester for a year).
 
Well, I don't claim actually to know anything. It just made me think that something may be different than normally I would expect. Perhaps it led somewhere useful. I think I flew south of England, once.
 
Hmmm.....make it 81

You have triangle ABC with sides 60, 81, 124: so easy to get angles.

You have triangle BCD with sides 81, 145, 130: again angles easy

Angle ABD = angle ABC - angle CBD ; ~39.089 degrees

Law of cosines makes AD = ~91.6071 ; ~91.0588 if 81.486 used

So that's one solution only...no?
 
Hello, yohanson77!

Okay, I'll go way out on a limb and guess that the decimal is a typo.
Here's my version of the problem:

A quadrilateral plot on the ground \(\displaystyle ABCD\) is surveyed for redevelopment.
It has boundary dimensions: \(\displaystyle AB\,=\,60.0,\:BC \,=\,81.0,\:CD\,=\,145.0\)
and has internal diagonal measurements of: \(\displaystyle CA\,=\,124.0\) and \(\displaystyle BD\,=\,130.0\)
all measured in meters.\ .Determine the length AD.
Code: 
            A       60        B
            * - - - - - - - - *
           /    *         *    \
          /         * *         \
         /        *     *        \ 81
        /     * 130     124 *   θ \
       /  *                     *  \ 
      * - - - - - - - - - - - - - - *
      D            145              C

Let \(\displaystyle \angle ACB\,=\,\theta\)

Consider \(\displaystyle \Delta ABC\).
Code: 
            A       60        B
            * - - - - - - - - *
                *              \
                    *           \
                    124 *        \ 81
                            *   θ \
                                *  \ 
                                    * C

Law of Cosines: \(\displaystyle \:\cos\theta \:=\:\frac{81^2\,+\,124^2\,-\,60^2}{2(81)(124)} \:=\:0.912833532\)

Hence: \(\displaystyle \:\theta \:=\:24.10009047^o\:\approx\:24.1^o\)


Consider \(\displaystyle \Delta BCD.\)
Code: 
                              B
                              *
                          *    \
                130   *         \
                  *              \ 81
              *                   \
          *                        \ 
      * - - - - - - - - - - - - - - *
      D            145              C

Law of Cosines: \(\displaystyle \:\cos C \;=\;\frac{81^2\,+\,145^2\,-\,130^2}{2(81)(145)} \:=\:0.454916986\)

Hence: \(\displaystyle \:C \:=\:62.94040691 \:\approx\:62.9^o\)

Then: \(\displaystyle \,\angle ACD \:=\:62.9\,-\,24.1 \:=\:38.8^o\)


Now consider \(\displaystyle \Delta ACD.\)
Code: 
             A
            *
           /    *
          /         * 124
         /              *
        /                   *
       /                 38.8°  *
      * - - - - - - - - - - - - - - *
      D            145              C

Law of Cosines: \(\displaystyle \:AD^2\;=\;124^2\,+\,145^2\,-\,2(124)(130)\cos38.8^o \;=\;8376006781\)

Therefore: \(\displaystyle \:AD \;=\;91.52052656 \;\approx\;91.5\text{ m}\)

 
Sorry for the long delay, internet has been off. I'm sorry to have caused all the confusion but it is my mistake, the comma is supposed to be a decimal. my sausage fingers must have hit it.

Sorry to all concerned, but many thanks for the replies. The last post especially made loads of sense, thank you again:)
 
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