Answer:
Hence, the azimuth, bearing, and length of the line connecting station Shore to station Rock are [tex]270^{\circ},90^{\circ}[/tex] and [tex]650.94[/tex] feet.
Given :
The shore station point [tex](X_1,Y_1)[/tex] in feet,
[tex](X_1,Y_1)=(2058.97, 6980.06)[/tex]
The Rock station point [tex](X_2,Y_2)[/tex] in feet
[tex](X_2,Y_2)=(1408.03,6980.06)[/tex]
From the figure
[tex]x=2058.97-1408.03=650.94[/tex] feet
[tex]y=6980.06-6980.06=0\\[/tex] feet
Length of the line [tex]L=\sqrt{x^2+y^2}[/tex]
[tex]\Rightarrow L=\sqrt{(650.94)^2+0}=650.94[/tex]
[tex]\Rightarrow L=650.94[/tex] feet
[tex]\because[/tex] [tex]\tan \theta=\frac{y}{x}=\frac{0}{x}=0[/tex]
[tex]\Rightarrow \theta=\tan^{-1}(0)=0[/tex]
Azimuth of line[tex]=270^{\circ}+\theta[/tex]
[tex]=270^{\circ}+0[/tex]
[tex]=270^{\circ}[/tex]
[tex]\therefore[/tex] Bearing [tex]=360^{\circ}-\text{Azimuth}[/tex]
[tex]=360^{\circ}-270^{\circ}[/tex]
[tex]=90^{\circ}[/tex]
