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Trigonometry: Height of Pole

  1. Jul 31, 2012 #1
    Having a fair bit of trouble with this one. Based on the answer I had thought to apply the Sine Rule but adding points to describe the lengths of elevation seems wrong, but I'm stuck on how else to approach this. Can anyone help, please?

    Many thanks.

    1. The problem statement, all variables and given/known data

    Q. A vertical flagpole stands on horizontal ground. The angle of elevation of the top of the pole from a certain point on the ground is [itex]\theta[/itex]. From a point on the ground 10m closer to the pole, the angle of elevation is [itex]\beta[/itex]. Show that the height of the pole is [itex]\frac{10sin\theta sin\beta}{sin(\beta-\theta)}[/itex].

    3. The attempt at a solution

    Points of pole: top c, bottom d.
    Point at elevation [itex]\theta[/itex] = a
    Point at elevation [itex]\beta[/itex] = b
    [itex]sin\theta=\frac{a}{sin\theta}{|ac|}[/itex] = [itex]|ac|=\frac{a}{sin\theta}[/itex]
    [itex]sin\beta=\frac{b}{|bc|}[/itex] = [itex]|bc|=\frac{b}{sin\beta}[/itex]
    [itex]|\angle acb|180^o-(sin\theta+90^o) = 180^o-sin\theta[/itex]

    I'd keep going at this point but I have a feeling I'm way off...
     
  2. jcsd
  3. Jul 31, 2012 #2

    eumyang

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    No, the Sine rule isn't used here.
    (I labeled |ab| as 10, |bd| as x, and |cd| as h.)
    Find tan θ in terms of 10, x and h. Find tan β in terms of x and h. Rewrite the latter equation so that x is by itself, and substitute into the former equation. Use trig identities to "simplify" to the answer.
     
  4. Jul 31, 2012 #3
    This is how things are looking for me now:
    a. [itex]tan\theta=\frac{h}{10+x}[/itex]
    b. [itex]tan\beta=\frac{h}{x}[/itex] = [itex](tan\beta)x=h[/itex] = [itex]x=\frac{h}{tan\beta}[/itex]
    Thus, b.->a. = [itex]tan\theta=\frac{h}{10+h/tan\beta}[/itex] = [itex]h=tan\theta(10+\frac{h}{tan\beta})[/itex] = [itex]h=\frac{10tan\theta+h(tan\theta)}{tan\beta}[/itex]

    I seem to be quite close to the answer, but something has gone wrong here. Can you help me out again, please?
     
  5. Jul 31, 2012 #4

    eumyang

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    From here:
    [itex]h=\tan \theta(10+\frac{h}{\tan \beta})[/itex],
    distribute the tan theta and move the fraction to the other side. You want to isolate the h, without having another h term on the other side.
     
    Last edited: Jul 31, 2012
  6. Aug 1, 2012 #5
    Ok, continuing on:

    [itex]h=\frac{10tan\theta+(h)tan\theta}{tan\beta}[/itex] = [itex]h-\frac{(h)tan\theta}{tan\beta}=\frac{10tan\theta}{tan\beta}[/itex] = [itex]\frac{(h)tan\beta-(h)tan\theta}{tan\beta}=\frac{10tan\theta}{tan \beta}[/itex] = [itex]h(tan\beta-tan\theta)=\frac{10tan\theta\cdot tan\beta}{tan\beta}[/itex] = [itex]h=\frac{10tan\theta\cdot tan\beta}{tan\beta(tan\beta-tan\theta)}[/itex]

    The common denominator, in the very last part of the attempt here, should probably be [itex]tan(\beta-\theta)[/itex], but I'm not sure how to factor that. Also, how do I switch from tan to sin? I'm nearly there! Can you help again, please?

    Thank you.
     
  7. Aug 1, 2012 #6

    eumyang

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    From here:
    [itex]h=\tan \theta(10+\frac{h}{\tan \beta})[/itex]
    ...to here is wrong. It should be
    [itex]h = 10\tan \theta + \frac{h\tan \theta}{\tan \beta}[/itex]
    Subtract the 2nd fraction (the one with the h in the numerator) from both sides. Factor out the h to get
    [itex]h( something ) = 10\tan \theta[/itex]
    Divide both sides by that (something).

    Also, can you please stop putting equal signs between each step? Instead, you can press "Enter" between steps.
     
  8. Aug 1, 2012 #7
    Ok, so I've made the correction and continued on from there. That just leaves me with the job of converting tan to sin. How is that achieved?

    a. [itex]tan\theta=\frac{h}{10+x}[/itex]
    b. [itex]tan\beta=\frac{h}{x}[/itex]
    = [itex](tan\beta)x=h[/itex]
    = [itex]x=\frac{h}{tan\beta}[/itex]
    Thus, b.->a. = [itex]tan\theta=\frac{h}{10+h/tan\beta}[/itex]
    = [itex]h=tan\theta(10+\frac{h}{tan\beta})[/itex]
    = [itex]h=\frac{10tan\theta tan\beta+h(tan\theta)}{tan\beta}[/itex]

    =[itex]h=\frac{10tan\theta tan\beta+(h)tan\theta}{tan\beta}[/itex]
    = [itex]h-\frac{(h)tan\theta}{tan\beta}=10tan\theta[/itex]
    = [itex]\frac{(h)tan\beta-(h)tan\theta}{tan\beta}=10tan\theta[/itex]
    = [itex]h(tan\beta-tan\theta)=10tan\theta\cdot tan\beta[/itex]
    = [itex]h=\frac{10tan\theta\cdot tan\beta}{tan\beta-tan\theta}[/itex]
     
  9. Aug 2, 2012 #8

    eumyang

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    Those equal signs you are putting at the beginning of each line: you are using them incorrectly. Either don't put them at the beginning of each line, or use \rightarrow within the itex tags.
    Using the trig identities, rewrite the expression in terms of sines and cosines. Now you have a complex fraction. Multiply the numerator and denominator by something to simplify the fraction.
     
  10. Aug 2, 2012 #9
    Ok, I have it now. Thank you.
     
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