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Basic Trig

  1. Mar 29, 2006 #1
    URGENT Trig question

    Okay I missed this lesson, so I dont really know what to do. I know it has to do with law of Sines/Law of Cosines. (but I think I can do all the parts involving that)

    1)First Decide how many triangles can be drawn based on the information (my edit: I can do this)
    2)then draw it or them as accuratly as possible (My edit: I can do the "it" part its the second half of the them part thats the problem)
    3)Solve for any missing side or angle in any triangle drawn. (I can do this when I get that far)
    4)Find the area of each triangle to the nearest integer (Thats easy)

    problems:


    1) In Tri. ABC m<A=30, a=7, b=8

    Okay, because b>a>bSinA I know that two triangles can be drawn from this information.

    I can draw and find all of the information (Including the area) for the first triangle. What I cant do is find how to draw the second triangle. I talked to my friend and she said it was really easy and all I had to do was to find the referance 30 in quad. 2, and that would be the obtuse angle of my triangle. However, I have no idea what to do beyond that. I think she must have left out a couple of steps. (It was a quick conversation)

    If I merely replace <A with 150 then no triangle can be drawn from the given information.

    Ok, I'm lost, any help?
     
    Last edited: Mar 29, 2006
  2. jcsd
  3. Mar 29, 2006 #2

    Hootenanny

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    Is it a requirement to draw these triangles? From what I can see it is a simple application of the sine rule.
     
  4. Mar 29, 2006 #3
    yes.

    Unforunatly, I have no idea what you mean by "simple apllication of the sine rule"
     
  5. Mar 29, 2006 #4

    Hootenanny

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    So, from what I understand you wish to draw right triangles 'inside' triangle ABC, in order to find the area of ABC?
     
  6. Mar 29, 2006 #5
    no, I need to draw two seperate and different triangles based on the information given. The area part is just a little added bonus thats easy.
     
  7. Mar 29, 2006 #6

    Hootenanny

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    Okay, i've sketched this triangle as follows: the base is b, the left slope is c and the other side is a opposite the 30 angle. Now you can draw two triangles;

    (1)With the side c, part of side b and a vertical side. You can use trig to find the values of these sides.
     
  8. Mar 29, 2006 #7
    I can find the first triangle. Its the second I cant find. And unless Im mistaken, (Which is entirly possible) YOu hvae only showed me how to find the first. Basicly, Im just really confused.
     
  9. Mar 29, 2006 #8

    Hootenanny

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    Right, ok so call the vertical side of the triangle h. Have you found h and the length of part of b (call this b')?
     
  10. Mar 29, 2006 #9
    Huh? Ik dont think so. all the info I have on the first tri is here.
    Myabe this will help- here is teh information I have for the first triangle.
    P.S. I was suppposed to round all answers to nearest tenth for length and nearest min for angles.
    <A=30
    a=7
    <B=34 deg 51min
    b=8
    <C=115 deg 9 min
    c= 127
    Area= 25

    edit: and Height is equal to 7.241586163)
     
    Last edited: Mar 29, 2006
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