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Homework Help: 3d word problem

  1. Apr 4, 2015 #1
    1. The problem statement, all variables and given/known data
    A vertical mast stands on the north bank of a river with straight parallel banks running from east to west. The angle of elevation of the top of the mast is α when measured from a point A on the south bank distant 3a to the east of the mast and β when measured from another point B on the south bank distant 5a to the west mast. prove that the height of the mast is

    2. Relevant equations

    3. The attempt at a solution
    Let the height of the mast be h.
    Let C be the base of the mast.
    Let D be the point at which the perpendicular from BA to C divides BA.

    BC = \frac{h}{tan β}\\\\
    CA = \frac{h}{tan α}\\\\
    By Pythagoras' theorem

    CD = \sqrt{(\frac{h}{tan β})^2 - (5a)^2)}\\\\
    CD = \sqrt{(\frac{h}{tan α})^2 - (3a)^2)}\\\\

    \sqrt{(\frac{h}{tan β})^2 - (5a)^2)} = \sqrt{(\frac{h}{tan α})^2 - (3a)^2)}\\\\\
    (\frac{h}{tan β})^2 - (\frac{h}{tan α})^2 = (5a)^2 - (3a)^2\\\\
    \frac{h^2 tan^2 α - h^2 tan^2 β }{(tan^2 β) (tan^2 α)} = 16a^2\\
    h = 4a\sqrt{\frac{(tan^2 β) (tan^2 α)}{tan^2 α - tan^2 β }}

    At this point I figure that either the question is floored or i've made a mistake. Usually it's the latter.
    Last edited: Apr 4, 2015
  2. jcsd
  3. Apr 4, 2015 #2


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    ... or you have to take this a bit further.

    What is ##\displaystyle\ \left(\frac{(tan^2 β) (tan^2 α)}{tan^2 α - tan^2 β }\right)^{-1}\ ## ?
  4. Apr 4, 2015 #3
    Ah yes, i'm kicking myself. Thanks for the nudge.
    Last edited: Apr 4, 2015
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