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October 1, 2014

October 1, 2014

Posted by **Mojo** on Saturday, May 21, 2011 at 6:14pm.

- Precalculus -
**MathMate**, Saturday, May 21, 2011 at 10:10pmLet H=horizontal distance to the pole

Vertical distance

=Htan(θ)

where θ is the angle of depression.

Thus

60' = Htan(θ2)-Htan(θ1)

=H(tan(18°)-tan(14°))

Solve for H to get

H = 60'/(tan(18°)-tan(14°))

=?

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