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Triangles Problem

  1. Apr 30, 2003 #1
    I found a neat problem on the web you can look at it here I've always had fun with these find the angle problems but my solutions on this one seemed a little bit weird. Anyone want to let me know what they got? Heres what I found angle BCA = 20 and angle DBC = 50 (the geometry sure doesn't look that way, but I can't find an error in the way I did it and everything seems to be self consistent with my answers...)
     
    Last edited by a moderator: Apr 30, 2003
  2. jcsd
  3. May 1, 2003 #2
    I got the same thing. I'ld forgotten how to get those last two angles because I haven't done anything like that for a few years but when I thorght about all of the infomation that I knew I got the same awnsers.
     
  4. May 1, 2003 #3
    climbhi, that's a nasty one
    I spent some time trying to solve it by the old 180° rule, but didn't succeed. Probably, trig is needed. Being too lazy, I just graphed it. The answer is probably
    <BCA = 60°, <DBC = 10°.
    Anybody know a nice analytical way to solve it?
     
    Last edited: May 1, 2003
  5. May 1, 2003 #4

    Hurkyl

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    Well, it's easy enough to prove they have to add up to 70. However, the reason you are having trouble is because, for all the triangles in the diagram, they both appear in the same triangles.

    So the solution is to create new triangles! My suggestion is to draw a line through C parallel to AD, and see what you can do.
     
  6. May 1, 2003 #5
    Yeah drawing new triangles was critical to getting the answers that I got. It's nice to see enslam got the same thing. The way I did it was to draw two new triangles branching off what was already there from which I was able to get a system of four equations with four unknowns, and had my calculator solve from there. The reason I'm uncertain is the way it looks on the page it sure doesn't look like angle DBC is greater then angle BCA.

    I don't know so much about linear algebra so maybe the system that I had wasn't fully determined (could very well be the wrong vocab word, I really don't know too much about solving systems with lin. alg.) which lead to there being an answer to the system which was not physically possible?

    I'm certain these equations are right, perhaps someone who knows more linear algebra then I do could look at them and tell me if what I suspected up above is right:
    x + y = 70
    x + b = 50
    a + b = 90
    y + a = 110
     
  7. May 1, 2003 #6
    Now I look at it agin the awnser I gave can only be true if DA is parallel to CB and DC is parallel to AB. In this case its not because CDA = 100 and DAB = 100, which means its impossible for DC to be parallel to AB. Thus my awnser was incorrect.
     
  8. May 1, 2003 #7
    Hmm well you and I got the same answers but I can't see why those answers are dependant on those criteria. So far I think I've checked every possible quadrilateral and triangle in the thing and the angle measures in them all sum up to 360 and 180 respectively, so the answers appear to be self consistent...
     
  9. May 1, 2003 #8

    Use the sine rule and the cosine rule to calculate the length between each of the lines. With this info you can find out what the able to find out what the missing angles are.

    arcnets was right. The awnser is <BCA = 60°, <DBC = 10°.

    What I was getting at with the parallel line is that when a parallel line is bisected with another line the angles are the same. I can't really explain it any better but if u draw two parallel lines and bisect then you'll see what I'm getting at. <ADB and DBC were the same but that can only occur when AB and DC are parallel (stated earlier). Because they weren't parallel then the angles couldn't be the same.
     
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