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Trigo/Geom. problem

  1. May 23, 2009 #1

    Mentallic

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    1. The problem statement, all variables and given/known data
    D is the top of a tower h metres high. A is the foot of the tower. The angles of depression of the points B and C from D are [itex]45+\theta[/itex] and [itex]45-\theta[/itex] respectively. Show that [itex]BC=2h.tan2\theta[/itex]

    http://img43.imageshack.us/img43/9195/trigo.th.jpg [Broken]

    3. The attempt at a solution

    [tex]\angle ADB=90-(45+\theta)=45-\theta[/tex]

    Therefore, [tex]tan(45-\theta)=\frac{AB}{h}[/tex]

    Then, [tex]AB=h.tan(45-\theta)[/tex]

    [tex]\angle CDA=90-(45-\theta)=45+\theta[/tex]

    Therefore, [tex]tan(45+\theta)=\frac{AB+CD}{h}[/tex]

    Then, [tex]AB=h.tan(45+\theta)-CD[/tex]

    So I equated [tex]h.tan(45-\theta)=h.tan(45+\theta)-CD[/tex]

    And finally, through expanding using the tangent sum formula and 'compressing' back into the double tangent formula, I resulted in [itex]BC=2h.tan2\theta[/itex]


    My problem is that this way to do it took far too long. There must be an easier way but I cannot seem to find it. I need your help :smile:
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 23, 2009 #2

    HallsofIvy

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    Are B and D assumed to be in a straight line with A?
     
  4. May 23, 2009 #3

    Mentallic

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    Yes. The diagram is simply a right-angled triangle with another line going from D to a point B on the opposite line of the triangle.

    EDIT: Re-reading your question Hallsofivy, I think I may need to post a diagram to clear things up.
     
  5. May 23, 2009 #4

    Mentallic

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  6. May 23, 2009 #5

    Mentallic

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    Bump. To re-iterate: I doubt such a question given to our class to answer would require over a page of algebraic manipulations, and this is why I'm thinking there is a much easier way to use that [itex]2\theta[/itex] in the diagram directly, and not obtain it through the tangent double angle formula.
     
  7. May 24, 2009 #6

    HallsofIvy

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    If the angle of depression to B is 45+[itex]\theta[/itex], then the angle DB makes with DA is 90- (45+ [itex]\theta-[/itex])= 45-[itex]\theta[/itex] and the angle DC makes with DA is 90- (45- [itex]\theta[/itex]= 45+[itex]\theta[/itex]. CA/h= tan(45+\theta) and BA/h= tan(45-[itex]\theta[/itex]) so CB= CA- BA= h(tan(45+[itex]\theta[/itex])- tan(45-[itex]\theta[/itex])). Now use the identities for tan(a+ b) and tan(a-b).
     
  8. May 24, 2009 #7

    tiny-tim

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    Hi Mentallic! :smile:

    (have a theta: θ :wink:)
    yeees :redface:

    having got your AB = h.tan(45º - θ),

    you never actually said AC = h.tan(45º + θ) …

    which gives you the answer much quicker! :smile:
     
  9. May 25, 2009 #8

    Mentallic

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    Oh thats much quicker! Thanks to both of you :smile:

    Still curious though, can this problem be solved without the trig identities? Such as using the [itex]2\theta[/itex] angle and finding its tangent etc.?
     
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