# Math Help - Bearing word problems help

1. ## Bearing word problems help

They are driving me crazy. Please help..

1.) A plane took off on a runway having a bearing of 75 degrees. After flying for 1.5 miles, the pilot turned 90degrees and headed toward northwest. After the plane goes 3 miles in this direction, what bearing should the control tower use to locate the aircraft?

2.) Two ships leave a pier at the same time. Ship A sails in the direction of 55degrees at 20 knots while ship b sails in the direction of 145 at 25 knots. Determine after traveling 3 hrs the distance between the ships. What is the bearing of ship A from ship B?

3.) A ship leaves a port and sails for 3 hrs on a course N78degrees E at 12 knots. Then the ship changes its course to 168degrees 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.

-----
Please help.. Thank you very much!

2. The first step is to draw a diagram (assuming you aren't using vectors yet). I have drawn it for the first one.

Then just start filling in the things you can deduce from the picture.

3. Hello, fayeorwhatsoever!

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