Hello, Helltech!

A ship is sailing due North. At a certain point the bearing of a lighthouse

that is 12 miles away is found to be N 39° E.

Later the bearing is determined to be N 44° E.

How far, to the nearest tenth of a mile, has the ship traveled? Code:

N o L
| * *
| * *
| * 5°*
| * *
| * *
| * *
| 44° * *
| * * 12
B o 136° *
| *
x | *
|39°*
| *
A o

The lighthouse is at $\displaystyle L.$

When the ship is at $\displaystyle A,\:\angle NAL = 39^o\text{, and }AL = 12$

When the ship is at $\displaystyle B,\:\angle NBL = 44^o\quad\Rightarrow\quad \angle LBA = 136^o$

We want: .$\displaystyle x \:=\:AB$

In $\displaystyle \Delta ABL,\;\angle BLA \:=\:180^o - 136^o - 39^o \:=\:5^o$

Law of Sines: . $\displaystyle \frac{x}{\sin5^o} \:=\:\frac{12}{\sin136^o} \quad\Rightarrow\quad x \:=\:\frac{12\sin5^o}{\sin136^o} \:=\:1.505587433$

Therefore, the ship traveled about 1.5 miles.