Solution:
A ship travels 6km south, and then 8km west.
The diagrammtic expression is shown below
To find θ, we will apply SOHCATOA
Given
![\begin{gathered} Opposite=8\text{ km} \\ Adjacent=6\text{ km} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/k0a83u6hv526xygbcfxttnpgss6umxgjro.png)
Applying the tan formula
![\tan\theta=(Adjacent)/(Hypotenuse)](https://img.qammunity.org/2023/formulas/mathematics/high-school/lk20cw0nb9gkk35b19xjv7epaejmgj6es0.png)
Substitute the values of the side lengths into the formula above
![\begin{gathered} \tan\theta=(8)/(6) \\ \tan\theta=1.3333 \\ \theta=\tan^(-1)(1.3333) \\ \theta=53.13\degree \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/9nqbmy1ivsu6k9zgag1x6wuk0aqmtv4393.png)
The bearing of the ship from the source to destination will be
![\begin{gathered} =\theta+180\degree \\ =53.13+180\degree=233.13\degree \\ =233\degree\text{ \lparen nearest degree\rparen} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/p5uyqjs6p8ug0p281iuwc7otykmqh94k82.png)
Hence, the answer is 233° (nearest degree)