Answer:33.66 km
Explanation:
Given
First sailboat travel [tex]40^{\circ}[/tex] south of east for 22 km to reach the buoy
Position vector of buoy [tex]r_1[/tex]
[tex]r_1=22cos40\hat{i}-22sin40\hat{j}[/tex]
Final Position of sailboat [tex]r_3[/tex]
[tex]r_3=15\hat{j}[/tex]
from Figure we can say that
[tex]r_1+r_2=r_3[/tex]
[tex]r_2=r_3-r_1[/tex]
[tex]r_2=15\hat{j}-22cos40\hat{i}+22sin40\hat{j}[/tex]
[tex]r_2=-22cos40\hat{i}+29.14\hat{j}[/tex]
[tex]|r_2|=33.66 km[/tex]
After reaching [tex]r_3[/tex]
[tex]tan\theta =\frac{22cos40}{29.14}[/tex]
[tex]tan\theta =\frac{16.85}{29.14}[/tex]
[tex]\theta =30.04^{\circ}east\ of\ south[/tex]