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Math Help - trigonomerty - modelling and problem solving

  1. #1
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    trigonomerty - modelling and problem solving

    Three points A, B and C are on a horizontal line such that AB = 70m and BC = 35m. The angles of elevation of the top of the tower are x, y and z, where tan x = 1/13, tan y = 1/15, and tan z = 1/20. The foot of the tower is at the same level as A, B and C. Find the height of the tower.

    I can draw the diagram and have 3 equations in 2 unkowns, can't go any further. Please help
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  2. #2
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    Re: trigonomerty - modelling and problem solving

    I have tried out the three equations. the value of h i got was 35 but on verification with third equation i got tan z = 1/16. Please verify the values.
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  3. #3
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    Re: trigonomerty - modelling and problem solving

    Can you explain further. Thanks
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  4. #4
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    Re: trigonomerty - modelling and problem solving

    Hello, elmidge!

    I agree with ibdutt: . \tan z must be \tfrac{1}{16}
    But then, we don't need point C or angle z.


    \text{Three points }A, B, C\text{ are on a horizontal line}
    . . \text{ such that }AB = 70m\text{ and }BC = 35m.

    \text{The angles of elevation of the top of the tower are }x, y, z,
    . . \text{ where: }\,\tan x = \tfrac{1}{13},\;\tan y = \tfrac{1}{15},\text{ and }\tan z = \tfrac{1}{16}

    \text{The foot of the tower is at the same level as }A, B\text{ and }C.

    \text{Find the height of the tower.}

    The tower is: h = PQ.
    Code:
        P *
          |** *
          | * *   *
          |  *  *     *
        h |   *   *       *
          |    *    *         *
          |     *     *           *
          |    x *    y *           z *
        Q * - - - * - - - * - - - - - - - *
            13h   A  70   B      35       C
    In \Delta PQA: \:\tan x = \frac{h}{QA} \quad\Rightarrow\quad \frac{1}{13} = \frac{h}{QA} \quad\Rightarrow\quad QA = 13h

    In \Delta PQB: \:\tan y = \frac{h}{13h+70} = \frac{1}{15} \quad\Rightarrow\quad 15h = 13h +70

    Therefore: . h \:=\:35\text{ m}
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  5. #5
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    Re: trigonomerty - modelling and problem solving

    You're both making the assumption that the base of the tower is on the same line as A, B and C.
    That needn't be the case, and if it isn't it's perfectly possible to arrive at an answer consistent with the given values.
    I have to go out now, but I can post a solution tomorrow unless either of you do so in the meanwhile.
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  6. #6
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    Re: trigonomerty - modelling and problem solving

    Let the tower be of height h with base at D, (not on the line ABC), and top E.
    Then,

    \frac{DE}{AD}=\frac{1}{13}, so AD = 13h, and similarly BD = 15h \text{ and } CD = 20h.

    We now have a triangle ACD, with AC =105, AD = 13h, CD = 20h, with a point

    B \text{ on } AC \text{ such that } AB = 70, BC = 35, \text{ and } BD = 15h.

    Here are two different methods for finding the value of h.

    (1) Using the cosine rule in the two triangles ABD and ACD.

    \cos A = \frac{169h^{2}+70^{2}-225h^{2}}{2.13h.70} \text{  from ABD,  and  }\cos A = \frac{169h^{2} + 105^{2}-400h^{2}}{2.13h.105} \text{  from ACD}.

    Tidy up and equate the two and find that h^{2}=25, \text{  so,  } h=5.

    (2) Using some co-ordinate geometry.

    Arrange for the line ABC to be along the x-axis with A at the origin.

    Take circles radii 13h, 15h and 20h at centres A, B and C respectively. The base of the tower D, will be at a common point of intersection.


    The circles have equations

    x^{2}+y^{2}= 169h^{2},\dots (1),

    (x-70)^{2}+y^{2}=225h^{2}, \dots (2)

    (x-105)^{2}+y^{2}=400h^{2}, \dots (3).

    Expand the two brackets and subtract (1) from (2), (1) from (3) yielding (4) and (5),

    -140x+4900=56h^{2}, \dots (4)

    -210x+11025 = 231h^{2} \dots (5).

    Now multiply (4) by 3, (5) by 2 and subtract one from the other to get rid of x.

    Tidy up and again find that h^{2}=25, \text{  so,  } h=5.
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    Re: trigonomerty - modelling and problem solving

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  8. #8
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    Re: trigonomerty - modelling and problem solving

    Thank you.
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