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Trigonometry problem

  1. Oct 7, 2016 #1
    1. The problem statement, all variables and given/known data


    I don't understand how to get CE.

    2. Relevant equations
    Trigonometric identities (maybe)

    3. The attempt at a solution

    First, I start with the cosine rule
    CB^2 = AC^2 + AB^2 - 2 AC AB cos α
    (CE/sin(β))^2 = (CE/sin(α))^2 + L^2 - 2 (CE/sin(α)) L cos α

    which is gonna be a quadratic equation and it's hard to obtain what CE is (using this method)
    Please help
  2. jcsd
  3. Oct 7, 2016 #2

    Simon Bridge

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    Science Advisor
    Homework Helper

    Try the usual way you derive the cosine rule - except solving for the altitude of the triangle instead of factoring it out.
  4. Oct 7, 2016 #3
    I think using sine rule to find a side first is a faster way
  5. Oct 7, 2016 #4
    I get this ##(\frac{1}{(sin^2(\beta))})CE^2+(\frac{2\ L\cos \alpha}{sin \ \alpha})CE-L^2=0##

    Then, use the quadratic formula to solve CE??
    But if I use quadratic formula, there will be two solutions, but there is only one CE.
    And I doubt if the solution obtained will be the same as in the photo I gave
  6. Oct 7, 2016 #5
    CB/ sin alpha = CA / sin beta

    CE = CA sin alpha
    CE = CB sin beta

    Then whatt??
  7. Oct 7, 2016 #6
    angle ACB = 180 - a -b
    you can try to use L/sin(180-a-b) = ??
  8. Oct 7, 2016 #7
    Angle ACB = 180 - (a + b)
    sin(180-(a+b)) = sin (a+b)

    L/sin(a+b) = CB/ sin (a) = CA / sin (b)

    How to make CE (the altitude) show up o_O?
  9. Oct 7, 2016 #8
    You succeeded to express CB (or CA) in terms of L , a and b. Then what is CB sin(b) ?
  10. Oct 7, 2016 #9

    L/sin(a+b) = CE/sin(a)Sin(b)
    CE = L sin(a) sin(b)/ sin(a+b)

    Thanks a lot for your help bro
  11. Oct 7, 2016 #10

    Ray Vickson

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    Science Advisor
    Homework Helper

    ##AE/CE = \cot(\alpha)## and ##BE/CE = \cot(\beta)##. Now add and manipulate.
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