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Math Help - Triangle Problem! Help

  1. #1
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    Triangle Problem! Help

    My teacher said that this was solvable. I have to figure out all the unknown sides and angles. Does anybody out there know how I would do this??? What are the first steps I need to do? I got as far as C=14
    Here is the triangle group:
    http://i711.photobucket.com/albums/w...gleProblem.jpg
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  2. #2
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    Quote Originally Posted by Leandra View Post
    My teacher said that this was solvable. I have to figure out all the unknown sides and angles. Does anybody out there know how I would do this??? What are the first steps I need to do? I got as far as C=14
    Here is the triangle group:
    http://i711.photobucket.com/albums/w...gleProblem.jpg

    You'll need to use several methods to solve each side and angle of the triangles. The first thing I'd recommend you do is Find each angle. We know that interior angles of a triangle all add up to 180. So, look at the triangle to the left. There are two angles given, 56 and 110 degrees = 166 degrees. Subtract 166 from 180 to get 14; angle C is 14 degrees.

    Since angle C is next to the other angle beside it, you can add them together when looking at the whole triangle (all three combined). So, the lower left corner of the large triangle is 34 degrees.

    From here you'll need to use simple trigonometric functions to find the lengths of the sides. You would do this the same way you'd solve a problem where you find the height of a tree (given one angle and the length of one side).

    Other than this, you may need to use the law of sines to find a few others. Law of sines is simple but takes a long time to explain, so I won't get into it. Also remember that the three angles in the middle will all equal 360 in the end, so if you know two of them, you can find the other by adding those two and subtracting from 360.

    Good luck
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  3. #3
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    I know the law of sines and cosines and that the middle angles add up to 360. But your suggestion is not working.
    I can't find any lengths of the sides with 34 degrees because I would need a right angle. I need at least two angles and a side don't I? Or an angle and 2 sides somewhere.
    Please help!
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  4. #4
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    Quote Originally Posted by Leandra View Post
    I know the law of sines and cosines and that the middle angles add up to 360. But your suggestion is not working.
    I can't find any lengths of the sides with 34 degrees because I would need a right angle. I need at least two angles and a side don't I? Or an angle and 2 sides somewhere.
    Please help!
    Hmm good point. It slipped my mind. I'm not sure, looking at the diagram I can't find any of the three triangles solvable. I'm thinking there is some rule that I'm forgetting though. Did your teacher give any more information on the problem? Also, I'm thinking you might have to split one of the triangles into two right triangles. I'm just not sure which one, where, or how. I remember doing this on problems before.
    Last edited by AlderDragon; May 14th 2009 at 11:05 PM.
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  5. #5
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    The teacher didn't give anymore info. All he said was that is was solvable and he sat down and did it at home the night before.

    Does the line z, if drawn to meet the line with the length 5 make a 90 degree angle? I think it does. Maybe something can be done with that? I'm running out of tylonol!
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  6. #6
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    Quote Originally Posted by Leandra View Post
    The teacher didn't give anymore info. All he said was that is was solvable and he sat down and did it at home the night before.

    Does the line z, if drawn to meet the line with the length 5 make a 90 degree angle? I think it does. Maybe something can be done with that? I'm running out of tylonol!
    Theres no way to know for sure unless you know angle D I'm not sure how much knowing that would help you though.
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  7. #7
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    Well here's my contribution. You could try to make a system of equations and solve them simultaneously:

    First solve the things that are relatively straight forward. But there's only one easy thing like that that I can spot (if someone else can spot more, then that might help with the problem I come to at the end of this attempt).

    In the small triangle on the left, angles in a triangle add up to 180 so: C+110+56=180\Rightarrow C=14

    Now here are the simultaneous equations:

    In the center, angles around a point add up to 360, so: A+B=360-110\Rightarrow A+B=250.

    The angles in the big composite triangle must add up to 180 so: D+E+56+14+20+70=180\Rightarrow D+E=20

    In the triangle at the bottom: B+E+20=180\Rightarrow B+E=160

    And in the triangle on the right: A+D+70=180\Rightarrow A+D=110

    Collecting the 4 relevant equations just for clarity:
    [1] A+B=250.
    [2] D+E=20
    [3] B+E=160
    [4] A+D=110

    So we have 4 unknowns and 4 equations. Which I thought should be solvable by using a series of substitutions to eliminate one of the letters from each equation in turn. But, whenever I try to solve it I keep coming out with 0=0 at the end. Does anyone know why this might happen? (Is it because the equations in the system are not linearly independent?) Don't worry, I'm sure someone can help with this.
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  8. #8
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    Quote Originally Posted by Leandra View Post
    My teacher said that this was solvable. I have to figure out all the unknown sides and angles. Does anybody out there know how I would do this??? What are the first steps I need to do? I got as far as C=14
    Here is the triangle group:
    http://i711.photobucket.com/albums/w...gleProblem.jpg
    Are you sure that the data given (the picture) is correct?

    It is not obvious, but from the information given the point in the middle of the triangle is in a different plane that the outer three vertices. Just a quick calculation indicates that the lower right 20(deg) angle will NEVER intersect the rightmost (E+70 angle) vertex.

    It appears as if you are looking down on a pyramid, with the A_B_110 point at the high point of the pyramid.

    A very rough (unchecked/unverified) calc seems to indicate that the 20 degree angle cannot be greater that 5.63 degrees.
    or
    the 70 degree must be less that 11.42 degrees.

    Just wondering.
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  9. #9
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    The triangle in not drawn in scale. My teacher is very lazy like that. He makes multiple problems with the same drawing just changing the measurements.
    If he is wrong or lying that the triangle is solvable then he will expect us to prove that it is not since he is mean.

    Is there a way to prove that it is an impossible set of triangles?

    The problem is due tomorrow!
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  10. #10
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    I think I am onto something!! Tell me what you guys think!!

    Left Triangle:
    C+110+56=180
    C=14

    Big Triangle:
    D+E+56+14+70+20=180
    D+E=20

    Bottom Triangle:
    B+E+20=180
    B+E=160

    Right Triangle:
    A+D+70=180
    A+D=110

    Center:
    A+B+110=360
    A+B=250

    D≠ 0 or 20 => 0 < D < 20
    E≠ 0 or 20 => 0 < E < 20
    A≠ 0 or 250 => 0 < A < 250
    B≠ 0 or 250 => 0 < B < 250

    Bottom Triangle:
    B(max) = 180-20-E(min) = 180-20-0= 160
    B(min) = 180-20-E(max) = 180-20-20= 140
    140<B<160 => B≠140 or 160

    Right Triangle:
    A(max) = 180-70-D(min) = 180-70-0= 90
    A(min) = 180-70-D(max) = 180-70-20= 70
    70<A<90 => A≠70 or 90

    Center:
    B(max) = 360-110-A(min) = 360-110-70= 180
    B(min) = 360-110-A(max)= 360-110-90= 160
    160<B<180 => B≠160 or 180

    This tells me that there is no solution for B since as stated above:
    140<B<160 B≠140 or 160
    160<B<180 B≠160 or 180


    Is my thinking right or wrong??
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  11. #11
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    Quote Originally Posted by Leandra View Post
    The triangle in not drawn in scale. My teacher is very lazy like that. He makes multiple problems with the same drawing just changing the measurements.
    If he is wrong or lying that the triangle is solvable then he will expect us to prove that it is not since he is mean.

    Is there a way to prove that it is an impossible set of triangles?

    The problem is due tomorrow!
    Here is an accurate diagram using the angles that were specified.


    The angle QSR is supposed to be 70. As you can see, it is much less than that. To be 70, it would have to lie on the circular arc from Q to R. Since that arc does not meet the line PS, the problem has no solution.
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