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Math Help - Three-Dimensional Problem?

  1. #1
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    Three-Dimensional Problem?

    Refer to 7(b):
    http://i45.tinypic.com/11smtr7.jpg

    Help, please? The answer given in the textbook states 57.5 degrees, but I can't seem to get it.
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  2. #2
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    Re: Three-Dimensional Problem?

    Quote Originally Posted by fActor View Post
    Refer to 7(b):
    http://i45.tinypic.com/11smtr7.jpg

    Help, please? The answer given in the textbook states 57.5 degrees, but I can't seem to get it.
    1. Triangle \Delta(ABP) is an isoscles triangle because |AP| = |BP|.

    2. Use Pythagorean theorem to determine the length of AP:

    The diagonal d of the base rectangle has the length 20, thus half of the diagonal (\frac12 d) has the length 10.

    |\overline{AP}|=\sqrt{5^2+10^2}=5\sqrt{5}

    3. Use Cosine rule to evaluate

    \cos(\angle(BAP))=\frac{(BP)^2-(AP)^2-(AB)^2}{-2 \cdot (AB) \cdot (BP)}=\frac 6{5\sqrt{5}}

    You'll get: \angle(BAP) = 57.544^\circ
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  3. #3
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    Re: Three-Dimensional Problem?

    A slightly different method: Set up a coordinate system with origin at A, positive x-axis from A to B, positive y-axis from A to D, and positive z-axis upward. Then the vector AB is <12, 0, 0> and AP is <6, 8, 5>. The magnitude of AB is 12, the magnitude of AP is 5\sqrt{5}, and the dot product of A and B is 6(12)= 72. We must have 6(12)= (12)(5\sqrt{5})cos(\theta) so that cos(\theta)= \frac{6}{5\sqrt{5}}.
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