91Ó°ÊÓ

Let us draw a figure

$$\begin{gathered} A=StThomas \\ B=WestIndies \end{gathered}$$

$AC=18×4=72miles$

Use law of cosines

$$\begin{gathered} C{B}^2=A{C}^2+B{C}^2-2\left(AC\right)\left(BC\right)\cos \left({15}^{∘}\right) \\ C{B}^2={72}^2+{200}^2-2\left(72\right)\left(200\right)\cos \left({15}^{∘}\right) \\ =5184+40000-28800\cos \left({15}^{∘}\right) \\ =17365.3262 \\ CB=\sqrt{17365.3262}=131.7776 \end{gathered}$$

Use law of sines to find angle

$$\begin{gathered} \frac{\sin {15}^{∘}}{131.7776}=\frac{\sin \mathrm{γ}}{200} \\ \sin \mathrm{γ}=200×\frac{\sin {15}^{∘}}{131.7776} \\ \mathrm{γ}={\sin }^{-1}\left(\frac{200\sin {15}^{∘}}{131.7776}\right) \\ \mathrm{γ}=23.{13}^{∘} \end{gathered}$$

Time to cover distance of 200 mile is $=\frac{200}{18}=11.11hrs$

Taking off course by ${15}^{∘}$took four hours to the crews to find that they were off course.

Now determine the hours it took to sailboat to return to the British West Indies

$131.7776×\frac{1hr}{18mi}=7.32hrs$

add the total hours it took taking off course $4+7.32=11.32hrs$

Subtract the original hours for taking the right path from the hours it took taking off course$11.32-11.11=0.21hrs$