Bearing and distance

[hardworker2153]

I'm not sure how to figure bearings and distance. Here's the problem:

A worker at "tower A" travels in the direction N54 degrees W for 1020 meters to get to "Tower B". From tower B" the worker travels in the direction S78 degrees W for 628 meters to get to tower T" Compute the bearing and distance to travel to "Tower T" include a sketch of the situation

 

[soroban]

Hello, hardworker2153!

A worker at Tower A travels in the direction N 54°W for 1020 meters to get to Tower B.
From Tower B, he travels in the direction S 78° W for 628 meters to get to Tower T.
Compute the bearing and distance to travel to Tower T. .Include a sketch of the situation

                        B                   P
                        * - - - - - - - - - +
                     *  : *                 :
                  * 78° :   *               :
           628 *        :     *             :
            *           :       *           :
         *              :    1020 *         :
    T * - - - - - - - - +           *       :
                        Q             * 54° :
                                        *   :
                                          * :
                                            * A

Going from A to B, the worker travelled: \(\displaystyle AP \,=\,1020\cdot\cos54^o \,=\,599.5\) m north
. . and: \(\displaystyle PB\,=\,1020\cdot\sin54^o \,\approx\,825.2\) m west.

Going from B to T, the worker travelled: \(\displaystyle BQ\,=\,628\cdot\cos78^o\,\approx\,369.1\) m south
. . and: \(\displaystyle QT\,=\,628\cdot\sin78^o\,\approx\,614.3\) m west.

Hence, point T is: \(\displaystyle 599.5\,-\,369.1\:=\:230.4\) m north of A
. . and: \(\displaystyle 825.2\,+\,614.3\:=\:1439.5\) m west of A.

Their positions look like this:

    T *
      |    *
      |         *
230.4 |              *
      |                   *
      |                 θ      *
      * - - - - - - - - - - - - - - *
                1439.5              A

We have: \(\displaystyle \,\tan\theta\,=\,\frac{230.4}{1439.5}\,=\,0.160066673\)
. . Hence: \(\displaystyle \,\theta\:\approx\:9^o\)

Therefore, the bearing is: \(\displaystyle \,\fbox{N\,81^o\,W}\)


The distance is given by Pythagorus:
. . \(\displaystyle AT\:=\:\sqrt{230.4^2\,+\,1439.5^2} \:\approx\:\fbox{1457.8\text{ m}}\)

 

[galactus]

\(\displaystyle \L\\1020cos(54)=599.54\)

\(\displaystyle \L\\1020sin(54)=825.20\)

628cos(78)=130.57<-----This one

\(\displaystyle \L\\628sin(78)=614.28\)


T is 599.54-130.57=468.97 m north of A.

T is 825.2+614.30=1439.5 m west of A.

Pythagoras: T is \(\displaystyle \L\\\sqrt{468.97^{2}+1439.5^{2}}=\fbox{1513.97}\)

m from A.

\(\displaystyle \L\\tan^{-1}(\frac{468.97}{1439.5})\approx{18} \;\ degrees\)

Therefore, the bearing from A to T is \(\displaystyle \L\\\fbox{N72^{\circ}W}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


You know another way to do this is to use azimuths. If we were surveying that is how it would be done.

Let's assign point A coordinates of (0,0)

The coordinates of B are:

1020cos(306)=599.54 North
1020sin(306)=-825.2 East

The negative means we are west of our starting point of (0,0).

628cos(258)=-130.57
628sin(258)=-614.28

Continuing from point B (traverse), we get as the coordinates of T:

599.54+(-130.57)=468.97
-825.2+(-614.28)=-1439.47

\(\displaystyle \L\\\tan^{-1}(\frac{468.97}{-1439.47})\approx{-18} \;\ degrees\)

360+(-18)=342 degrees as an azimuth or \(\displaystyle \L\\\fbox{N72^{\circ}W}\), as before.

The distance is 1513.97 m, as before.