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Math Help - Distances

  1. #1
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    Distances

    I don't understand this:

    The plan shows the roads between towns A, B and C. Find the distance:
    a) from A to BC;
    b) from B to AC;
    c) from C to AB.
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  2. #2
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    from A to BC probably means the shortest distance from point A to some point on BC.
    Last edited by mr fantastic; December 25th 2010 at 06:05 PM. Reason: Restored deleted post. User banned.
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  3. #3
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    And the shortest distance to a line occurs at a right angle to that line.
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  4. #4
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    They give such answers: a) 15 km, b) 7 and 1/17 km (that's one number), c) 8 km.
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  5. #5
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    The first thing you will have to do is to use the cosine rule to evaluate each angle.
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  6. #6
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    It's an exercise from an 8th grade book, where trigonometry functions (sin, cos, tg) are two years away, that's why I posted it not into trigonometry section. You can't use triangle similarity at this stage as well. This exercise is given at the end of the "Pythagoras theorem" chapter, where usually mixed exercises are, from earlier (not necessarily about the theorem). And I personally can't see how you can apply Pythagoras theorem here.
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  7. #7
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    You apply the Pythatorean theorem to show that the angle at B is a right angle!
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  8. #8
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    I noticed that, but what does it give me? But I deduced that the fact that there's no small square indicating that B is a right angle, they want us to solve it differently...
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  9. #9
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    Hello, Evaldas!

    I'll do part (b).


    \text{The plan shows the roads between towns }A, B, C.
    \text{Find the distance: (b) from B to AC.}
    Code:
                        B
                        o
                      * |*
                    *   | *
             15   *     |  *  8
                *       |h  *
              *         |    *
            *           |     *
          *             |      *
      A o * * * * * * * o * * * o C
        :     17-x      D   x   :

    We have \Delta ABC with: . AB = 15,\;BC = 8,\;AC = 17

    Draw BD \perp AC.\;\text{ Let }h = BD.

    Let x = DC,\;17-x = AD.

    In right triangle BDC\!:\;h^2 + x^2 \:=\:8^2 \quad\Rightarrow\quad h^2 \:=\:64 - x^2 .[1]

    In right triangle BDA\!:\;h^2 + (17-x)^2 \:=\:15^2 .[2]


    Substitute [1] into [2]: . (64-x^2) + (17-x)^2 \:=\:225

    . . which simplifies to: . 34x \:=\:128 \quad\Rightarrow\quad x \:=\:\frac{64}{17}


    Substitute into [1]: . h^2 \:=\:64 - \left(\frac{64}{17}\right)^2 \:=\:\frac{14,\!400}{289}

    . . Therefore: . h \:=\:\sqrt{\frac{14,\!400}{289}} \:=\:\frac{120}{17} \;=\;7\frac{1}{7}

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  10. #10
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    Thank you Soroban. I get it now.
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  11. #11
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    distances

    Soroban's method does not require ABC to be a right triangle. If you notice that 17^2 = 15^2+ 8^2 ABC is aright triangle
    Area = 4x15 = 60 km^2
    60 =a1 x17/2 a1=120/17
    60 =a2 x8/2 a2=120/8
    60 = a3 x15/2 a3 = 120/15 where a's are altitudes to coressponding sides


    bjh
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