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Sin cosine rule

  1. May 30, 2008 #1
    soine cosine rule

    I have a question for you, I came across this question while revising for my exam on monday if anyone can answer this I'll be very impressed.

    Two different trangles ABC have AB=5cm, AC=3.2cm and angle ABC=35degrees
    calculate the difference between their areas.
     
  2. jcsd
  3. May 30, 2008 #2

    berkeman

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    Staff: Mentor

    Welcome to the PF, seboastien. I've moved your post from the General Math forum to the Homework Help forums. An important rule here is that homework/coursework (/revising/studying) questions must be posted here in Homework Help.

    Also, we do not give out answers to these types of questions. We do provide tutorial help, to help you to figure out the problem on your own. We need to see some of your own work, though, before we can help you.

    So, what are your ideas on how to approach this geometry problem? Why would the two triangles have different areas if they have two equal sides and an equal inclusive angle?
     
  4. May 30, 2008 #3

    symbolipoint

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    Which is the OTHER triangle? You want just the difference between a triangle and itself?
     
  5. May 30, 2008 #4
    who said that the sides were the same length? AB=5 and AC=3.2, who said they were the same?
    and to symbolipoint, that is the whole question which is why it makes no sense
     
  6. May 30, 2008 #5

    berkeman

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    It sounds to us like you are saying that both triangles have AB=5, and both have AC=3.2, and both have the same inclusive angle. That sounds like the triangles are equal. What are we missing?
     
  7. May 30, 2008 #6
    the angle of BAC and the side BC, are saying that they are describing two seprate triangles?
     
  8. May 30, 2008 #7
    What is the original question?
     
  9. May 30, 2008 #8
    two different triangles ABC have AB=5cm AC=3.2cm and angle ABC=35 degrees
    calculate the difference between the two sides
     
  10. May 30, 2008 #9
    it still doesn't make any sense, if the triangles are seprate then their area is exactly the same surely, using the sine rule you can find the angle C, then you could find A as you know the two other angles meaning there's only one possible answer for the area
     
  11. May 30, 2008 #10
    Angle ABC isn't the inclusive angle. Given the two sides you are given and the angle you are given, how many possible triangles are there? Can you draw it or them? Hint: Try both obtuse and acute triangles. Then, ask yourself what information you would need to find the area of each.
     
  12. May 30, 2008 #11

    Dick

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    Draw the segment AB. Now though point B draw a line that makes an angle of 35 degrees to the segment. The only requirement for point C is that it lies on the line and is 3.2cm from point A. So make a circle around point A of radius 3.2cm. The circle hits the line in two points, see?
     
  13. May 30, 2008 #12
    Hi seboastien...It does make sense if the angle is at B
    Draw a line AB of length 5cm, then from point A construct a circle of radius 3.2cm, Then draw a line from B inclined at 35degs from AB and see what you get.

    Sorry Dick/Tedjn...didn't see your reply as I posted this
     
  14. May 30, 2008 #13
    what do you mean inclusive angle
     
  15. May 30, 2008 #14
    i understand what your saying but i still appear to be getting the wrong answer
     
  16. May 30, 2008 #15
    according to the text book te answer is 4.07 cm sqrd but i'm getting 2.somthing
     
  17. May 30, 2008 #16
    1. The problem statement, all variables and given/known data
    two different triangles ABC where AB=5cm, AC=3.2cm, and angle ABC is 35 degrees, calculate the difference between the two areas


    2. Relevant equations
    SinA/a=SinB/b
    Area=0.5abSinC


    3. The attempt at a solution
    drawn line AB then a line at B 35 degrees to AB. then drawn a circle of centre A with a radius 3.2cm, I now have two triangles, one with both sides 3.2cm and the angle between 81.4
    the other triangle with sides 1.8cm and 5.5cm and the angle between being 35
    using Area=0.5abSinC I have found areas of both triangles and then I have found the difference between them, and I am getting the wrong answer
     
  18. May 30, 2008 #17

    Dick

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    How can anyone tell unless you post your incorrect solution?
     
  19. May 30, 2008 #18
    if you read my new "thread" as it were you will find my working
     
  20. May 30, 2008 #19
    it's half ten in the evenong here, and I should probably get some kip. thanks anyway
     
  21. May 30, 2008 #20

    Dick

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    I'm not finding any 81.4 angles. I would suggest you drop a perpendicular from A to the line BC and find it's length. Then you can just work with right triangles. Makes things easier for me, anyway. BTW the difference area you want to find is just the area of the isosceles triangle, isn't it? You don't really have to do two triangles.
     
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