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Trig Stuff Helpp!

  1. Feb 10, 2005 #1
  2. jcsd
  3. Feb 10, 2005 #2

    Zurtex

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    Ugh, I hate not answering these maths questions in the help forum because they are jumped on by like 20 mathematicians because not enough maths questions are posted up. But I'll be the one to ask:

    What have you done so far?
     
  4. Feb 10, 2005 #3

    dextercioby

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    Do you know the addition formula...?

    Daniel.
     
  5. Feb 10, 2005 #4
    yes I do, that sin a + b = sin a cos b + cos a sin b..

    I know that one, don't know where to begin.
     
  6. Feb 10, 2005 #5
    HELPP!! please...

    jkotha
     
  7. Feb 10, 2005 #6

    dextercioby

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    Then set b=a and see what happens.

    Daniel.
     
  8. Feb 10, 2005 #7
  9. Feb 10, 2005 #8

    dextercioby

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    I couldn't open the wabpage.Could you type in the problem exactly,to see what it asks for...?

    Daniel.
     
  10. Feb 10, 2005 #9
  11. Feb 10, 2005 #10
    I have to use the triangle given, And I need to do it in 2 different ways.
     
  12. Feb 10, 2005 #11
    Please Helpp!
     
  13. Feb 10, 2005 #12

    Zurtex

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    Please stop posting so much and show the working you have attempted to do to work out the area of the triangle.

    dextercioby you should be able to right click and save as, once on your desktop open it with Acrobat Reader.
     
    Last edited: Feb 10, 2005
  14. Feb 10, 2005 #13

    dextercioby

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    What's the area of a triangle as a function of two consecutive sides and the angle between them?

    Daniel.
     
  15. Feb 10, 2005 #14
    Area of a triangle

    I know that the area of a triangle is 1/2 base * height. This would give the area of one of the triangles. We double this to find the area of the whole triangle.

    In reply to your post would it be 1/2(height) * C * sin (x)
    is that correct?
     
    Last edited: Feb 10, 2005
  16. Feb 10, 2005 #15
    also in the picture there is a 2nd triangle, would this right angle triangle have the same size as the two placed next to each other because of some math rule? (i have no idea which one it may be)

    that would make c equal to 2a

    thanks for the help

    j
     
  17. Feb 10, 2005 #16

    dextercioby

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    Yes,it would be correct.Now use the fact that the total area of the big triangle can be written as twice the area of the smaller (rectangle) one.
    It should come out easily.

    Daniel.
     
  18. Feb 10, 2005 #17
    k I understand the area of the triangle part just fine, but what confuses me is proving it to the identity with the sines and cosines.

    the height is not given explicitly in the picture, its blank.. so i used pythagorean theorem to find the height and it is (sq root of) \/ c^2 - a^2 or if i used c = 2a then it would be \/ 3a^2

    But im not sure if this leads me anywhere sooo ill approach at another angle.

    Since it asks us to find 2 ways, then the 2nd way i think i just found is to turn the triangle on its side (that part doesnt really matter but easier to place in a circle) and then say that cosine is the bottom plane (formerly height) and sine is a, so that would mean 1/2cos(x)*sin(x) is the area of the triangle.
    And if i look the identity sin(2x)=2sin(x)cos(x) and what i got for the triangle area 1/2cos(x)sin(x), then doubling the latter would give me 2sin(x)cos(x) i think and that would make it equal to sin(2x), therefore giving me the area of the triangles.

    is this 2nd part i just wrote right? (if u can even understand it lol) then that would be half way done.

    thanks
     
    Last edited: Feb 10, 2005
  19. Feb 11, 2005 #18

    dextercioby

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    The area of the big triangle is definitely:
    [tex] S=\frac{c^{2}}{2}\sin 2x [/tex]

    As i said,it is double the are of the rightangle triangle,which is (denoting the height of the big triangle by h) denoted by "s" and is
    [tex] s=\frac{c}{2} h \sin x [/tex]

    And now the "h" is (from the same triangle):
    [tex] h=c \cos x [/tex]

    And so,the "s" becomes
    [tex]s=\frac{c^{2}}{2}\sin x \ \cos x [/tex]

    Now double "s" and set it equal to S.

    Daniel.
     
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