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Math Help - trigonometry with triangle

  1. #1
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    trigonometry with triangle

    The pt T is the top of a vertical lighthouse.Boats are moored at pts A,B andC.The foot O of the lighthouse A,B and C are all at sea level.The boat A is 85m from the ptO and the angle of elevation of T from A is 24degress .The boats A and B are due east of lighthouse and the distance AB is 40m.The boat atC is due south of the boat at A and the angle of elevation of T from C is 18 degrees.
    Caculate
    1)The distance OT in m to the nearest m,
    2)Thje distance OB
    3)The angle of elevation of the pt T from the boat at Bin degrees to the nearest degree.
    Calculate also in m to the nearest metre
    4)the distance CT
    5)The distance OC
    6)The distance AC
    7)Calculate the bearing of the boat at B from the boat at C in degrees to the nearest degree.
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, NIKITA!

    Did you make some sketches?


    The point T is the top of a vertical lighthouse.
    Boats are moored at points A, B\text{ and }C.
    The foot O of the lighthouse, and A,B,C are all at sea level.

    Boat A is 85m from O; the angle of elevation of T from A is 24.
    Boats A\text{ and }B are due east of lighthouse and distance AB is 40m.
    Boat C is due south of boat A; the angle of elevation of T from C is 18.

    Caculate:

    1) The distance OT to the nearest meter.
    Code:
        T *
          | * *
          |   *   *
          |     *     *
          |       *       *
          |     24 *         *
          * - - - - - * - - - - - *
          O    85     A    40     B
    In right triangle TOA\!:\;\;\tan24^o \:=\:\frac{OT}{85} \quad\Rightarrow\quad OT \:=\:85\tan24^o

    Therefore: .  OT \:=\:37.844... \:\approx\: \boxed{38\text{ m}}



    2) The distance OB
    OB \;=\;OA + AB \;=\;85 + 40 \;=\;\boxed{125\text{ m}}


    3) The angle of elevation of T from B to the nearest degree.
    In \Delta TOB\!:\;\;\tan B \:=\:\frac{38}{125} \:=\:0.304

    Therefore: . B \:=\:\tan^{-1}(0.304) \:=\:16.909... \:\approx\:\boxed{17^o}



    4) The distance CT
    Code:
        T *
          |   *
          |       *
       38 |           *
          |               *
          |               18 *
          * - - - - - - - - - - - *
          O                       C

    In \Delta TOC\!:\;\;\sin18^o \:=\:\frac{38}{CT}\quad\Rightarrow\quad CT \:=\:\frac{38}{\sin18^o}

    Therefore: . CT \:=\:122.970... \:\approx\:\boxed{123}\text{ m}



    5) The distance OC

    In \Delta TOC\!:\;\;\tan18^o \:=\:\frac{38}{OC} \quad\Rightarrow\quad OC \:=\:\frac{38}{\tan18^o}

    Therefore: . OC \:=\:116.951... \:\approx\:\boxed{117\text{ m}}



    6) The distance AC
    Looking straight down at the ground, we have:
    Code:
                    85
        O * - - - - - - - - * A
             *              |
                *           |
               117 *        |
                      *     |
                         *  |
                            * C
    From Pythagorus: . AC^2 + 85^2 \:=\:117^2

    Therefore: . AC \:=\:\sqrt{117^2-85^2} \:=\:80.399... \:\approx\:\boxed{80\text{ m}}



    7) Calculate the bearing of B from C, nearest degree.
    Again, looking straight down at the ground . . .
    Code:
                  40
        A * - - - - - - - - * B
          |              *
          |           *
       80 |        *
          | θ   *
          |  *
          *
          C
    We have: . \tan\theta \:=\:\frac{40}{80}\quad\Rightarrow\quad \theta \:=\:\tan^{-1}(0.5)

    Therefore: . \theta \:=\:26.565... \:\approx\:\boxed{27^o}

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