Oblique Triangle Diagram Problem

ronjerby

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Mar 9, 2019
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I came across with a sample problem about Oblique Triangle in the internet that goes like this:
(Refer to: http://www2.clarku.edu/faculty/djoyce/trig/oblique.html)

P and Q are two inaccessible points. To find the distance between them, a point A is taken in QP produced, and a line AB 1200 feet long is measured making the angle PAB = 26° 35'. The angle ABP = 48° 12' and ABQ = 106° 42'. How long is PQ?

The hint for this problem, which is written in the same website is this: To find PQ, first find AP and AQ. You can find AP using the law of sines on triangle ABP, and you can find AQ using the law of sines on triangle ABQ.

The answer is (which is also written in the site): 651.9 feet.

I tried to solve the problem but I can't get past drawing the figure.

Can anyone help in providing the appropriate diagram for this problem?
 

MarkFL

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I think your diagram could look something like this:

11310
 

ksdhart2

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It sounds to me like you might be having difficulties drawing a diagram with an angle of exactly 26 degrees, 35 minutes. Well, the good news is you don't need to do that at all. Pick any two points P and Q and draw the line connecting them. Pick any point on the line PQ and label it A. Then draw any point B any arbitrary angle and distance away from A. Draw lines connecting all four points, such that you form two triangles.

Then label the diagram with the given angles and length. The diagram doesn't need to be exact, to scale, or even remotely accurate. The diagram serves purely as a visual aid to help you determine the trigonometry you need to do to find the real answer.
 

ronjerby

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I think your diagram could look something like this:

View attachment 11310
Thanks, MarkFL! Your diagram is flawless :D

I was confused where to place point A to accommodate the given values.

Again, thanks for sharing your thoughts, MarkFL :)
 

Denis

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Mark, I don't understand your diagram.
AB is given as 1200', PQ = ~652' (per given answer).
But your AB is shown much shorter than your PQ.

My diagram was approximately:
Code:
                B



Q                    P                              A
angleABP=~48, angleBAP=~26, so angleAPB=~106

Am I drunk? :)
 

Jomo

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Otis

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… Am I drunk?
Only on hockey. :p

It took me too long to draw my diagram to post (wasn't sure what "in QP produced" meant), but it ended up close to yours. As ksdhart2 noted, however, diagrams for these types of exercises don't really need to match the measurements because we don't rely on the accuracy of the diagram to set up the ratios; the diagram serves as a way to organize the info. (If you didn't have the given answer, you wouldn't have known that AB is almost twice as long as PQ.)

Getting the correct values into the Law-of-Sines ratios is priority one.

By the way, my answer differs from the given answer by 1/10th of a foot: 651.8 (rounded from 651.7539). Did I goof?

😎
 

MarkFL

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Mark, I don't understand your diagram.
AB is given as 1200', PQ = ~652' (per given answer).
But your AB is shown much shorter than your PQ.

My diagram was approximately:
Code:
                B



Q                    P                              A
angleABP=~48, angleBAP=~26, so angleAPB=~106

Am I drunk? :)
Yeah, I only looked at some of the angles, not the distances. I think your placement of the points is much better. :)
 

ronjerby

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Thanks for the input ksdhart2!

I was confused when I assumed that A is in between P and Q. It is impossible for Angle ABQ to have an angle of 100 degrees. I have no idea where to put A.
 

Otis

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… I was confused when I assumed that A is in between P and Q …
Me too!

Later, I realized that the phrase "in QP produced" means "on the line QP extended", as shown by MarkFL (post #2) and Denis (post #5). Note the web page said QP, not PQ, so point A is to the right of P. Cheers

😎
 

ronjerby

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Wow, thanks for the help, guys - MarkFL, ksdhart2, Jomo and Otis!

I really appreciated your inputs.
 
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