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Dodgy trig question

  1. Jan 8, 2006 #1
    in the diagram (attached), ABCD is a quadrilateral where AD parallel to BC. it is given that AB=9, BC=6, CA=5 and CD=15.
    show that cos BCA=-1/3 and hence find the value of sin BCA

    i can do the 'show that' bit but i don't know how to find sinBCA. any help??

    also when evaluating log15 + log20 - log12 (all to the base 5) do you do the addition to multiplication conversion first or the subtraction to division first?

    Attached Files:

  2. jcsd
  3. Jan 8, 2006 #2
    i cant see the triangle yet... beu about the logs... what does it matter what order you do it?
  4. Jan 8, 2006 #3


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    re: logs

    HINT 1: Multiplication is commutative

    HINT 2: Division is the mulitiplicative inverse.
  5. Jan 8, 2006 #4


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    For the first part.

    You know that [tex]\theta =\cos^{-1}(-1/3)[/tex] where [tex]\theta = B\hat CA[/tex]

    [tex]\mbox{Let } \phi = \pi /2 - \theta[/tex]
    [tex]\cos\phi = -\cos\theta = -(-1/3) = 1/3[/tex]
    [tex]\sin\phi = \sin\theta[/tex]

    Draw a small right-angled triangle, with one of the angles as [tex]\phi[/tex].
    Mark the adjacent side as 1, and mark the hypotenuse as [tex]3[/tex].
    So, what is the length of the opposite side?
    What is the value of [tex]\sin\phi[/tex] ?
    What is the value of [tex]\sin\theta[/tex] ?
  6. Jan 8, 2006 #5

    i didn't really follow that.
    i don't get this bit: [tex]\cos\phi = -\cos\theta = -(-1/3) = 1/3[/tex]
    cos i don't think we've done stuff like that, at least we haven't been taught it, i can't find it on the syllabus and the exam is in 1 week. is there another way??

    also, that post about the logs with all the long words in it, i didn't understand that either. i'm only 17. in english, possibly??
  7. Jan 8, 2006 #6
    hehe, comutative means you can rearange the operators in what order youd like and youd get the same answer...

    so if multiplication is comutative a*b*c=c*a*b and if division by b is just multiplication by 1/b then division is also comutative... so theres no point in asking what to do first: add the logs or substruct them...
  8. Jan 8, 2006 #7


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    You do know that sin2x+ cos2x= 1 don't you?
  9. Jan 8, 2006 #8
    hmmm... i wonder where i went wrong here:
    I. [tex]|CB|*|BA|cos(CBA)+|BA|*|AC|cos(BAC)=9=45cos(CBA)+54cos(BAC)[/tex]
    II. [tex]|BC|*|CA|cos(BCA)+|BA|*|AC|cos(BAC)=6=30cos(BCA)+54cos(BAC)[/tex]
    III. [tex]|BC|*|CA|cos(BCA)+|CB|*|BA|cos(CBA)=5=30cos(BCA)+45cos(CBA)[/tex]
    from I-II-III i get [tex]9-6-5=-60cos(BCA)[/tex]... which means [tex]cos(BCA)=\frac{1}{30}[/tex]...
    hmm... you said you proved its [tex]\frac{-1}{3}[/tex]?
  10. Jan 8, 2006 #9


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    Oh, my gawd!! :cry:
  11. Jan 8, 2006 #10


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    You use the cosine rule on the triangle ABC.
  12. Jan 9, 2006 #11
    heh, i call it adding the two prejections of the other sides on one of the sides and saying it equals the whole side, but you can call it the "cosine rule"
    anyway, it should give the right result... but i got [tex]cos(BCA)=\frac{1}{30}[/tex] while in the question they mentioned [tex]cos(BCA)=\frac{-1}{3}[/tex]
  13. Jan 9, 2006 #12
    heh, oops :blushing:
    to find the projectio of |BC| for example on |CA| you only need |BC|cos(BCA)... thats what ive done wrong, i multiplied the two vectors instead of just projecting them... what a silly mistake, guess thats because it was really late at night when i did it... and the floor was in a slope too :biggrin:
  14. Jan 10, 2006 #13
    all that stuff with the lines in it looks really confusing...i just used the cosine rule...

    anyway thanks for all the help i got it in the end
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