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Math Help - bearings

  1. #1
    Junior Member
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    bearings

    I don't need solutions, I just need someone to help me visualize these questions so I can set up the problems.

    1. A pilot wishes to fly an airfield S20šE of his present position. If the average airspeed of the plane is 520 km/h and the wind is from N80šE at 46 km/h, what direction does he need to steer?

    2. A destroyer detects a submarine 8 nautical miles due east traveling northeast at 20 knots. If the destroyer has a top speed of 30 knots, at what heading should it travel to intercept the submarine?
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, ty2391!

    1. A pilot wishes to fly an airfield S20šE of his present position. If the average
    airspeed of the plane is 520 km/h and the wind is from N80šE at 46 km/h,
    what direction does he need to steer?
    Code:
                      O
        W - - - - - - * - - - - - - - - - - - E
              10° *    *  *  θ 
              * 46      *     * 520
        B *              *        *
              *           *a          *
                  *        *              * A
                      *     *         *
                          *  *    *
                              *
                              P
    \angle EOA = \theta, \quad |OA| = 520, \quad \overrightarrow{OA} \:=\:\langle 520\cos\theta,\:\text{-}520\sin\theta\rangle

    \angle WOB = 10^o,\quad |OB| = 46, \quad \overrightarrow{OB} \:=\:\langle\text{-}46\cos10^o,\:\text{-}46\sin10^o\rangle

    \angle EOP = 70^o, \quad |OP| = a, \quad \overrightarrow{OP} \:=\:\langle a\cos70^o,\:\text{-}a\sin70^o\rangle


    We want: . \overrightarrow{OA} + \overrightarrow{OB} \:=\:\overrightarrow{OP}

    . . \begin{array}{cccccccc}520\cos\theta - 46\cos10^o &=& a\cos70^o & \Rightarrow & a &=&\dfrac{520\cos\theta - 46\cos10^o}{\cos70^o} & {\color{blue}[1]} \\ \\[-3mm] \text{-}520\sin\theta - 46\sin10^o  &=&\text{-}a\sin70^o & \Rightarrow & a &=&\dfrac{520\sin\theta + 46\sin10^o}{\sin70^o} & {\color{blue}[2]}\end{array}


    Equate [1] and [2]: . \frac{520\cos\theta - 46\cos10^o}{\cos70^o} \:=\:\frac{520\sin\theta + 46\sin10^o}{\sin70^o}

    . . 520\sin70^o\cos\theta - 46\sin70^o\cos10^o \:=\:520\cos70^o\sin\theta + 46\cos70^o\sin10^o

    . . . 520\sin70^o\cos\theta - 520\cos70^o\sin\theta \;=\;46\sin70^o\cos10^o + 46\sin10^o\cos70^o

    . . . . 520(\sin70^o\cos\theta - \cos70^o\sin\theta) \;=\;46(\sin70^o\cos10^o + \sin10^o\cos70^o)

    . . . . . . . . . . . . . 520\sin(70^o-\theta) \;=\;46\sin(70^o+10^o)


    And you can solve for \theta.

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