# Bearing problem

• Dec 5th 2010, 10:17 AM
Brndo4u
Bearing problem
I do not need this problem solved. Im just not sure how the drawing of this problem should look.
If someone could show me how it should look that would be greatly appreciated.

The bearing of a lighthouse from a ship was found to be N 37�� E. After the ship sailed 2.5 mi due south, the new bearing was N 25�� E. Find the distance between the ship and the lighthouse at each location.
• Dec 5th 2010, 10:32 AM
skeeter
Quote:

Originally Posted by Brndo4u
I do not need this problem solved. Im just not sure how the drawing of this problem should look.
If someone could show me how it should look that would be greatly appreciated.

Distance Between a Ship and a Lighthouse ship was found to be N 37􏰂 E. After the ship sailed 2.5 mi due south, the new bearing was N 25􏰂 E. Find the distance between the ship and the lighthouse at each location.
The bearing of a lighthouse from a

apologize for the "mouse" tracks ...
• Dec 5th 2010, 11:30 AM
Brndo4u
Where does the N 37 E go in the drawing. I edited the problem it was worded wrong the first time.
• Dec 5th 2010, 12:40 PM
Soroban
Hello, Brndo4u!

Quote:

$\displaystyle \text{The bearing of a lighthouse from a ship was found to be }N\,37^o\,E.$
$\displaystyle \text{After the ship sailed 2.5 mi due south, the new bearing was }N\,25^o\,E.$
$\displaystyle \text{Find the distance between the ship and the lighthouse at each location.}$
Code:

    N                      ♠ L     :                  * *     :              *12o*     :          *    *     : 37o  *      *     :  *        *   A ♥ 143o      *     :        *     :      * 2.5 : 25o *     :  *     : *   B ♥

The lighthouse is at $\displaystyle \,L.$

When the shp is at $\displaystyle \,A,\:\angle N\!AL = 37^o.$

The ship sails 2.5 miles to $\displaystyle \,B\!:\:AB = 2.5;\;\angle NBL = 25^o$

We see that $\displaystyle \angle LAB = 143^o,\,\text{ hence: }\:\angle ALB = 12^o$

In $\displaystyle \Delta ALB$, use the Law of Sines:

. . $\displaystyle \displaystyle \frac{AL}{\sin25^o} \:=\:\frac{2.5}{\sin12^o} \quad\Rightarrow\quad AL \:\approx\:5.08\text{ miles.}$

. . $\displaystyle \displaystyle \frac{BL}{\sin143^o} \:=\:\frac{2.5}{\sin12^o} \quad\Rightarrow\quad BL \:\approx\:7.24\text{ miles.}$