bearings

ty2391 · #1 $Old$ May 13th, 2008, 07:41 PM

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 S20ºE of his present position. If the average airspeed of the plane is 520 km/h and the wind is from N80ºE 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?

Soroban · #2 $Old$ May 13th, 2008, 10:30 PM

Hello, ty2391!

Quote:

1. A pilot wishes to fly an airfield S20ºE of his present position. If the average
airspeed of the plane is 520 km/h and the wind is from N80ºE 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 $\theta.$