# Math Help - Word bearing Problem

1. ## Word bearing Problem

A ship is sailing due north. At a certain point the bearing of a lighthouse is N $48.6$E and the distance is 12.5. After a while, the captain notices that the bearing of the lighthouse is now S $42.9$ E. How far did the ship travel between the two observations of the lighthouse.

What's the distance?

2. Hello, qbkr21!

A ship is sailing due north.
At a certain point the bearing of a lighthouse is $N\,48.6\,E$ and the distance is 12.5 (miles?)
After a while, the captain notices that the bearing of the lighthouse is now $S\,42.9\,E$.
How far did the ship travel between the two observations of the lighthouse?
Code:
    B *
| *
|   *
|42.9°*
|       *
c|         *
|      88.5°* C
|        *
|48.6°* b=12.5
|  *
A *

At $A$ the ship takes a bearing of the lighthouse at $C$
. . and finds angle $A = 48.6^o$ and $b = AC = 12.5$ miles.

Later at point $B$, they find that angle $B = 42.9^o$.
. . Hence, angle $C\:=\:180^o - 48.6^o - 42.9^o \:=\:88.5^o$

To find side $c$, use the Law of Sines: . $\frac{c}{\sin C} = \frac{b}{\sin B}\quad\Rightarrow\quad c \:=\:\frac{b\sin C}{\sin B}$

So we have: . $c \;= \;\frac{12.5\sin88.5^o}{\sin42.9^o} \;=\;18.35659392$

Therefore, the distance is about $18.4$ miles.

3. ## Re:

Thanks Soroban you are incredible at math. I will learn from this.