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Soroban · #3 November 25th, 2008, 06:41 AM

Hello, fayeorwhatsoever!

Quote:

3) A ship leaves a port and sails for 3 hrs on a course N78°E at 12 knots.
Then the ship changes its course to 168° and sails for 5 hours at 10 knots.
After 8 hours, how far is the ship from the port?
Determine the bearing of the ship from the port.

Code:

      N       M
      :       :
      :       :
      :     A o 168°
      :      *  * 
      :     * 90° *
      :    *        *
      :78°* 36        *  50
      :  *              *
      : *                 *
      :*                    *
    P o                       *
            *                   *
                  *               * 
                        *           * 
                              *       * 
                                    *   *
                                          o B

The ship starts at port $P$ . . $\angle NPA = 78^o$
It sails 36 nautical miles to $A$ .

It changes course to $\angle MAB = 168^o$
. . and sails 50 nautical miles to $B$ .
Note that $\angle PAB = 90^o.$

Pythagorus: . $PB^2 \:=\:36^2 + 50^2 \:=\:3796 \quad\Rightarrow\quad PB \:=\:\sqrt{3796} \:\approx\:61.6$

The ship is about 61.6 nautical miles from the port.

In right triangle $PAB\!:\;\;\tan(P) \:=\:\tfrac{50}{36} \quad\Rightarrow\quad P \:=\:\tan^{-1}\left(\tfrac{50}{36}\right) \:\approx\:54.2^o$

The bearing is: . $\angle NPB \;=\;78^o + 54.2^o \;=\;132.2^o$

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